# Chemistry Stoichiometry Part 2

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 Author: Henri93 ID: 267767 Filename: Chemistry Stoichiometry Part 2 Updated: 2014-03-24 20:14:50 Tags: Chemistry Stoichiometry Part Folders: Description: Chemistry Stoichiometry Part 2 Show Answers:

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1. Mass from Volume
Calculate the mass of N2 needed to produce 43.2 liters of N2O4 when N2 combines with an excess of O2 at STP.
• Given: MN2 = ?  VN2O4 = 43.2L
• Equation: Mu
• Formula Mass: N 2*14.007 = 28.014
• Work:  5.40*101g of N2
2. Volume from Mass
Calculate the volume of N2 needed to produce 95.1 g of N2O5 when N2 combines with an excess of O2 at STP.
• Given: vN2 = ? mN2O5 = 95.1g
• Equation: Vu
• Formula mass: N 2*14.007 + O 5*15.999 = 108.009
• Work: 1.97 * 101 L of N2
3. Volume from Volume
Calculate the volume of O2 needed to produce 74.1 liters N2O when N2 combines with an excess of N2 at STP.
• given: vN2O = 74.1L, vO = ?
• equation: Vu =
• Work: 74.1 * 1/2 = 3.71 * 101L
4. Mols from Mols
Calculate the number of mols of N2 needed to produce 5.25 mols of ammonia when N2 combines with an excess of H2.
• Given: nN2 = ?, nNH3 = 5.25
• equation:
• Work: Nu = 5.25*1/2 = 2.63 * 100 mols
5. Mass from Mols
Calculate the mass of N2 needed to produce 4.23 mols of NO2 when N2 combines with an excess of O2.
• Given: mN2 = ?, nNO2 = 4.23
• equation:
• Formula Mass: N 2*14.007 = 28.014
• Work: 4.23*1/2 *28.014 = 5.928*101g of N2
6. Mass from Mass
Calculate the mass of FeS needed to produce 284g of FeCl2 when FeS combines with an excess of HCl.
• Given: mFeS = ?, mFeCl2 = 284g
• Equation:
• Work: 284/126.751 *1 *87.911 = 1.97*102g
7. 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: mAg = ?, mAgCl = 150g, PY = 63g
• Equation:
• 150/143.321 * 2/2 * 107.868 = 1.1*102g
• Percent Yield Equation: PY = Actual/theoretical
• Work: 63/110
• Percent Yield: 57.3%

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