Chemistry Stoichiometry Part 2
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Mass from Volume
Calculate the mass of N_{2} needed to produce 43.2 liters of N_{2}O_{4} when N_{2} combines with an excess of O2 at STP.
 Given: ^{M}N_{2} = ? ^{V}N_{2}O_{4} = 43.2L
 Equation: M_{u} =
 Formula Mass: N 2*14.007 = 28.014
 Work: 5.40*10^{1}g of N_{2}

Volume from Mass
Calculate the volume of N_{2} needed to produce 95.1 g of N_{2}O_{5} when N_{2} combines with an excess of O_{2} at STP.
 Given:^{ v}N_{2} = ? ^{m}N_{2}O_{5} = 95.1g
 Equation: V_{u} =
 Formula mass: N 2*14.007 + O 5*15.999 = 108.009
 Work: 1.97 * 10^{1} L of N_{2}

Volume from Volume
Calculate the volume of O_{2} needed to produce 74.1 liters N_{2}O when N_{2} combines with an excess of N_{2} at STP.
 given: ^{v}N_{2}O = 74.1L, ^{v}O = ?
 equation: Vu =
 Work: 74.1 * 1/2 = 3.71 * 10^{1}L

Mols from Mols
Calculate the number of mols of N_{2} needed to produce 5.25 mols of ammonia when N_{2} combines with an excess of H_{2}.
 Given: ^{n}N2 = ?, ^{n}NH3 = 5.25
 equation:
 Work: Nu = 5.25*1/2 = 2.63 * 10^{0} mols

Mass from Mols
Calculate the mass of N_{2} needed to produce 4.23 mols of NO_{2} when N_{2} combines with an excess of O_{2.}
 Given: ^{m}N_{2} = ?, ^{n}NO_{2} = 4.23
 equation:
 Formula Mass: N 2*14.007 = 28.014
 Work: 4.23*1/2 *28.014 = 5.928*10^{1}g of N_{2}

Mass from Mass
Calculate the mass of FeS needed to produce 284g of FeCl_{2} when FeS combines with an excess of HCl.
 Given: ^{m}FeS = ?, ^{m}FeCl_{2} = 284g
 Equation:
 Work: 284/126.751 *1 *87.911 = 1.97*10^{2}g

Percent Yield
Calculate the mass of Ag produced when 150g of AgCl are reacted with an excess of Zn. if the lab reaction only produced 63g of Ag what is the percent yield?
 given: ^{m}Ag = ?, ^{m}AgCl = 150g, PY = 63g
 Equation:
 150/143.321 * 2/2 * 107.868 = 1.1*10^{2}g
 Percent Yield Equation: PY = Actual/theoretical
 Work: 63/110
 Percent Yield: 57.3%