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Stoichiometry Stoichiometry uses the bridge Stoichiometry uses the bridge to solve problems Stoichiometry uses the bridge to solve problems To use the bridge To use the bridge To use the bridge there will be a given conversion that must be solved 30 in = To use the bridge there will be a given conversion that must be solved ft 30 in = To use the bridge there will be a given conversion that must be solved and conversion factors for you to use ft 30 in = To use the bridge there will be a given conversion that must be solved and conversion factors for you to use ft 12 in = 1 ft 30 in = To use the bridge there will be a given conversion that must be solved and conversion factors for you to use ft 12 in = 1 ft 3 ft = 1 yd 30 in = To use the bridge there will be a given conversion that must be solved and conversion factors for you to use ft 5280 ft = 1 mi 12 in = 1 ft 3 ft = 1 yd Conversion to solve ? 30 in = ft Conversion factors ? 5280 ft = 1 mi 12 in = 1 ft 3 ft = 1 yd 30 in = The number and unit is written in the top left ft 5280 ft = 1 mi 12 in = 1 ft 3 ft = 1 yd 30 in = The number and unit is written in the top left 30 in ft 5280 ft = 1 mi 12 in = 1 ft 3 ft = 1 yd Where is the number and unit written? 1 1 2 3 4 2 4 3 30 in = You have to choose which conversion to use 30 in ft 5280 ft = 1 mi 12 in = 1 ft 3 ft = 1 yd 30 in = You have to choose which conversion to use 30 in ft 5280 ft = 1 mi 12 in = 1 ft 3 ft = 1 yd 30 in = You have to choose which conversion to use and which way to put it in the bridge 30 in ft 5280 ft = 1 mi 12 in = 1 ft 3 ft = 1 yd 30 in = You have to choose which conversion to use and which way to put it in the bridge 30 in ft 12 in 1 ft 5280 ft = 1 mi 12 in = 1 ft 3 ft = 1 yd 30 in = You have to choose which conversion to use and which way to put it in the bridge These 2 units must match 30 in ft 12 in 1 ft 5280 ft = 1 mi 12 in = 1 ft 3 ft = 1 yd 30 in 12 in ? 1 ft ? 12 in = 1 ft 27 yd 1 yd ? 3 ft ? 1 yd = 3 ft 30 in = Then multiply all the top numbers 30 in ft 12 in 1 ft 5280 ft = 1 mi 12 in = 1 ft 3 ft = 1 yd 30 in = Then multiply all the top numbers 30 30 in ft 12 in 1 ft 5280 ft = 1 mi 12 in = 1 ft 3 ft = 1 yd 30 in = Then multiply all the top numbers 30x1 30 in ft 12 in 1 ft 5280 ft = 1 mi 12 in = 1 ft 3 ft = 1 yd 30 in = Then multiply all the top numbers 30x1 30 in ft and divide by the bottom numbers 12 in 1 ft 5280 ft = 1 mi 12 in = 1 ft 3 ft = 1 yd 30 in = Then multiply all the top numbers 30x1÷12 30 in ft and divide by the bottom numbers 12 in 1 ft 5280 ft = 1 mi 12 in = 1 ft 3 ft = 1 yd 30 in = Then multiply all the top numbers 30x1÷12 = 2.5 30 in ft and divide by the bottom numbers 12 in 1 ft 5280 ft = 1 mi 12 in = 1 ft 3 ft = 1 yd Solve 15 rev 1 rev 3 in 1 ft 12 in = The conversion factors in stoichiometry The conversion factors in stoichiometry come from several places. The conversion factors in stoichiometry When converting using grams come from several places. The conversion factors in stoichiometry When converting using grams come from several places. The conversion factors in stoichiometry When converting using grams use the periodic table come from several places. The conversion factors in stoichiometry When converting using grams use the periodic table come from several places. The conversion factors in stoichiometry When converting using grams to find the conversion. use the periodic table come from several places. The conversion factors in stoichiometry When converting using grams to find the conversion. use the periodic table come from several places. = To find the conversion factor for Palladium To find the conversion factor for Palladium To find the conversion factor for Palladium use the average atomic mass To find the conversion factor for Palladium use the average atomic mass To find the conversion factor for Palladium use the average atomic mass and write the conversion factor To find the conversion factor for Palladium use the average atomic mass and write the conversion factor 1 To find the conversion factor for Palladium use the average atomic mass and write the conversion factor 1 mol To find the conversion factor for Palladium use the average atomic mass and write the conversion factor 1 mol Pd To find the conversion factor for Palladium use the average atomic mass and write the conversion factor 1 mol Pd = To find the conversion factor for Palladium use the average atomic mass and write the conversion factor 1 mol Pd = 106.42 To find the conversion factor for Palladium use the average atomic mass and write the conversion factor 1 mol Pd = 106.42 g To find the conversion factor for Palladium use the average atomic mass and write the conversion factor 1 mol Pd = 106.42 g Pd 1 mol Pd = 1 mol Al = 1 mol He = 1 mol Fe = g Pd ? g He ? g Fe ? g Al ? To find the conversion factor To find the conversion factor for converting between substances To find the conversion factor for converting between substances you need a balanced chemical equation To find the conversion factor for converting between substances you need a balanced chemical equation 6Li + N2 → 2Li3N To find the conversion factor If you need to convert between Li and N for converting between substances you need a balanced chemical equation 6Li + N2 → 2Li3N To find the conversion factor If you need to convert between Li and N for converting between substances then use the numbers in the equation you need a balanced chemical equation 6Li + N2 → 2Li3N To find the conversion factor If you need to convert between Li and N for converting between substances then use the numbers in the equation you need a balanced chemical equation 6Li + N2 → 2Li3N To find the conversion factor If you need to convert between Li and N for converting between substances then use the numbers in the equation you need a balanced chemical equation 6Li + N2 → 2Li3N To find the conversion factor If you need to convert between Li and N for converting between substances then use the numbers in the equation you need a balanced chemical equation 6Li + N2 → 2Li3N 6 To find the conversion factor If you need to convert between Li and N for converting between substances then use the numbers in the equation you need a balanced chemical equation 6Li + N2 → 2Li3N 6 mol To find the conversion factor If you need to convert between Li and N for converting between substances then use the numbers in the equation you need a balanced chemical equation 6Li + N2 → 2Li3N 6 mol Li To find the conversion factor If you need to convert between Li and N for converting between substances then use the numbers in the equation you need a balanced chemical equation 6Li + N2 → 2Li3N 6 mol Li = To find the conversion factor If you need to convert between Li and N for converting between substances then use the numbers in the equation you need a balanced chemical equation 6Li + N2 → 2Li3N 6 mol Li = 1 To find the conversion factor If you need to convert between Li and N for converting between substances then use the numbers in the equation you need a balanced chemical equation 6Li + N2 → 2Li3N 6 mol Li = 1 mol To find the conversion factor If you need to convert between Li and N for converting between substances then use the numbers in the equation you need a balanced chemical equation 6Li + N2 → 2Li3N 6 mol Li = 1 mol N2 To find the conversion factor If you need to convert between Li and N for converting between substances then use the numbers in the equation you need a balanced chemical equation 6Li + N2 → 2Li3N 6 mol Li = 1 mol N2 with the unit moles To find the conversion factor If you need to convert between Li and N for converting between substances then use the numbers in the equation you need a balanced chemical equation 6Li + N2 → 2Li3N 6 mol Li = 1 mol N2 with the unit moles To find the conversion factor If you need to convert between Li and N for converting between substances then use the numbers in the equation you need a balanced chemical equation 6Li + N2 → 2Li3N 6 mol Li = 1 mol N2 with the unit moles mol Br2 = mol Al = mol Al = 2Al + 3Br2 → 2AlBr3 mol AlBr3 mol Br2 mol AlBr3 To solve this problem: To solve this problem: 27 g H2 = g O2 2H2 + O2 → 2H2O You must write a conversion for each unit To solve this problem: 27 g H2 = g O2 2H2 + O2 → 2H2O You must write a conversion for each unit To solve this problem: that is NOT moles 27 g H2 = g O2 2H2 + O2 → 2H2O You must write a conversion for each unit To solve this problem: that is NOT moles 27 g H2 = g O2 2H2 + O2 → 2H2O 1 mol H2 = 2.02 g H2 You must write a conversion for each unit To solve this problem: that is NOT moles 27 g H2 = g O2 2H2 + O2 → 2H2O 1 mol H2 = 2.02 g H2 1 mol O2 = 32 g O2 You must write a conversion for each unit To solve this problem: then find the conversion beween the substances using that is NOT moles 27 g H2 = g O2 2H2 + O2 → 2H2O 1 mol H2 = 2.02 g H2 1 mol O2 = 32 g O2 You must write a conversion for each unit To solve this problem: then find the conversion beween the substances using the balanced chemical equation that is NOT moles 27 g H2 = g O2 2H2 + O2 → 2H2O 1 mol H2 = 2.02 g H2 1 mol O2 = 32 g O2 You must write a conversion for each unit To solve this problem: then find the conversion beween the substances using the balanced chemical equation that is NOT moles 27 g H2 = g O2 2H2 + O2 → 2H2O 1 mol H2 = 2.02 g H2 1 mol O2 = 32 g O2 You must write a conversion for each unit To solve this problem: then find the conversion beween the substances using the balanced chemical equation that is NOT moles 27 g H2 = g O2 2H2 + O2 → 2H2O 1 mol H2 = 2.02 g H2 1 mol O2 = 32 g O2 2 mol H2 You must write a conversion for each unit To solve this problem: then find the conversion beween the substances using the balanced chemical equation that is NOT moles 27 g H2 = g O2 2H2 + O2 → 2H2O 1 mol H2 = 2.02 g H2 1 mol O2 = 32 g O2 2 mol H2 = 1 mol O2 To solve this problem: Then use the bridge to convert 27 g H2 = g O2 2H2 + O2 → 2H2O 1 mol H2 = 2.02 g H2 1 mol O2 = 32 g O2 2 mol H2 = 1 mol O2 To solve this problem: Then use the bridge to convert 27 g H2 = g O2 2H2 + O2 → 2H2O 1 mol H2 = 2.02 g H2 1 mol O2 = 32 g O2 2 mol H2 = 1 mol O2 27 g H2 To solve this problem: Then use the bridge to convert 27 g H2 = g O2 2H2 + O2 → 2H2O 1 mol H2 = 2.02 g H2 1 mol O2 = 32 g O2 2 mol H2 = 1 mol O2 27 g H2 To solve this problem: Then use the bridge to convert 27 g H2 = g O2 1 mol H2 2.02 g H2 2H2 + O2 → 2H2O 1 mol H2 = 2.02 g H2 1 mol O2 = 32 g O2 2 mol H2 = 1 mol O2 27 g H2 To solve this problem: Then use the bridge to convert 27 g H2 = g O2 1 mol H2 2.02 g H2 2H2 + O2 → 2H2O 2 mol H2 1mol O2 1 mol H2 = 2.02 g H2 1 mol O2 = 32 g O2 2 mol H2 = 1 mol O2 27 g H2 To solve this problem: Then use the bridge to convert 27 g H2 = g O2 1 mol H2 2.02 g H2 2H2 + O2 → 2H2O 2 mol H2 1mol O2 1 mol O2 32 g O2 1 mol H2 = 2.02 g H2 1 mol O2 = 32 g O2 2 mol H2 = 1 mol O2 27 g H2 To solve this problem: Then use the bridge to convert 27 g H2 = g O2 1 mol H2 2.02 g H2 2H2 + O2 → 2H2O 27 2 mol H2 1mol O2 1 mol O2 32 g O2 1 mol H2 = 2.02 g H2 1 mol O2 = 32 g O2 2 mol H2 = 1 mol O2 27 g H2 To solve this problem: Then use the bridge to convert 27 g H2 = g O2 1 mol H2 2.02 g H2 2H2 + O2 → 2H2O 27 ÷ 2.02 2 mol H2 1mol O2 1 mol O2 32 g O2 1 mol H2 = 2.02 g H2 1 mol O2 = 32 g O2 2 mol H2 = 1 mol O2 27 g H2 To solve this problem: Then use the bridge to convert 27 g H2 = g O2 1 mol H2 2.02 g H2 2H2 + O2 → 2H2O 27 ÷ 2.02 ÷ 2 2 mol H2 1mol O2 1 mol O2 32 g O2 1 mol H2 = 2.02 g H2 1 mol O2 = 32 g O2 2 mol H2 = 1 mol O2 27 g H2 To solve this problem: Then use the bridge to convert 27 g H2 = g O2 1 mol H2 2.02 g H2 2H2 + O2 → 2H2O 27 ÷ 2.02 ÷ 2 × 32 2 mol H2 1mol O2 1 mol O2 32 g O2 1 mol H2 = 2.02 g H2 1 mol O2 = 32 g O2 2 mol H2 = 1 mol O2 27 g H2 To solve this problem: Then use the bridge to convert 27 g H2 = g O2 1 mol H2 2.02 g H2 2H2 + O2 → 2H2O 27 ÷ 2.02 ÷ 2 × 32= 213.9 2 mol H2 1mol O2 1 mol O2 32 g O2 1 mol H2 = 2.02 g H2 1 mol O2 = 32 g O2 2 mol H2 = 1 mol O2 |