Gas Stoichiometry 求教

2013-10-17 8:55 am
1.CaO is produced by heating CaCO3. Calculate the volume of CO2 produced at STP from decomposition of 152 g of CaCO3 according to the reaction:

CaCO3 -----> CaO + CO2



2.How many grams of calcium carbonate will you heat in order to obtain 2.00 L of carbon dioxide in a reaction under STP conditions?

CaCO3 -----> CaO + CO2



3.How many liters of oxygen gas at 20.0 °C and 2.00 atm can be prepared by thermal decomposition of 3.00g of KClO3?

2 KClO3 (s) -----> 2KCl (s)+ 3O2(g)



4.What volume of NH3 gas will be produced from 100.0 g of N2 gas and 100.0 g of H2 gas in the following reaction at STP?
N2 (g) + 3H2 (g) ------> 2NH3 (g)

回答 (1)

2013-10-17 9:52 am
✔ 最佳答案
Molar volume of a gas is 22.4 Liter at STP.

1.CaO is produced by heating CaCO3. Calculate the volume of CO2 produced at STP from decomposition of 152 g of CaCO3 according to the reaction:

CaCO3 -----> CaO + CO2

No. of mole of CaCO₃ = 152/(40 + 12 + 16X3) = 1.52
No. of mole of CO₂ producted = 1.52
Vol. of CO₂ = 1.52X22.4 = 34.0 L

2.How many grams of calcium carbonate will you heat in order to obtain 2.00 L of carbon dioxide in a reaction under STP conditions?

CaCO3 -----> CaO + CO2

No. of mole of CO₂ = 2.00/22.4 = 0.0893
No. of mole of CaCO₃ required = 0.0893
Mass of CaCO₃ = (0.0893)(40 + 12 + 16X3) = 8.93 g

3.How many liters of oxygen gas at 20.0 °C and 2.00 atm can be prepared by thermal decomposition of 3.00g of KClO3?

2 KClO3 (s) -----> 2KCl (s)+ 3O2(g)

No. of mole of KClO₃ = 3.00/(39+35.5+16X3) = 0.0245
No. of mole of O₂ = (0.0245)(3/2) = 0.0368
PV =nRT
(2.00)(V) = (0.0368)(0.08206)(273+20)
V = 0.442 L

4.What volume of NH3 gas will be produced from 100.0 g of N2 gas and 100.0 g of H2 gas in the following reaction at STP?
N2 (g) + 3H2 (g) ------> 2NH3 (g)

No. of mole of N₂ = 100/(14X2) = 3.571
No. of mole of H₂ = 100/(1X2) = 50

No.of mole of NH₃ that can be formed = 2X3.571 = 7.142

Vol. of NH₃ = 7.142X22.4 = 160.0 L


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