What is the minimum oxygen volume needed at STP for the complete combustion of 0.500 mol of this gas?

2016-11-02 2:53 am

更新1:

Very stuck on this problem. I understand the PV=nRT equation and that density=mass/volume so far Here is the complete question Chemical analysis of a gas shows that it contains one carbon atom for every one hydrogen atom. The density of this gas is 1.161 g/liter at STP. What is the minimum oxygen volume needed at STP for the complete combustion of 0.500 mol of this gas?

更新2:

I am still confused after what to do when I find the molar mass. I currently have 72.128 g/mol, however the empirical formula CH does not have any multiples that equal that molar mass.

回答 (1)

micatkie

2016-11-02 3:06 am

For the gas :
Pressure, P = 1 atm
Temperature, T = 273 K
Gas constant, R = 0.08206 L atm / (mol K)
Density, d = 1.161 g/L
Molar mass, M = ? g/mol

PV = nRT and n = m/M
Then, PV = (m/M)RT
PM = (m/V)RT
PM = dRT
Hence, M = dRT/P

Molar mass of the gas, M = 1.161 × 0.08206 × 273 / 1 g/mol = 26.0 g/mol

Empirical formula of the gas = CH
Let (CH)n be the molecular formula of the gas.

Molar mass of the gas :
n (12.0 + 1.0) = 26.0
n = 2

Hence, molecular formula of the gas = C₂H₂

Combustion of the gas :
2C₂H₂(g) + 5O₂(g) → 4CO₂(g) + 2H₂O(g)
OR: Mole ratio C₂H₂ : O₂ = 2 : 5

No. of moles of C₂H₂ burned = 0.500 mol
No. of moles of O₂ needed = (0.500 mol) × (5/2) = 1.25 mol

Molar volume of gas at STP = 22.4 L/mol
Volume of O₂ needed at STP = (1.25 mol) × (22.4 L/mol) = 28.0 L

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