Question

1. given the equation:

2 c₂h₆(g) + 7 o₂(g) → 4 co₂(g) + 6 h₂o(g)
determine the number of liters of co₂ formed at stp when 240.0 g of c₂h₆ is burned in excess oxygen gas. (molar mass c₂h₆ = 30.1g/mol)
a. 255 l
b. 437 l
c. 104 l
d. 712 l
e. 358 l

Explanation:

Step1: Calculate moles of \(C_2H_6\)

Use the formula \(n=\frac{m}{M}\), where \(m = 240.0\ g\) (mass of \(C_2H_6\)) and \(M=30.1\ g/mol\) (molar mass of \(C_2H_6\)).
\(n_{C_2H_6}=\frac{240.0\ g}{30.1\ g/mol}\approx7.97\ mol\)

Step2: Use mole - ratio from the balanced equation

From \(2C_2H_6(g)+7O_2(g)\to4CO_2(g)+6H_2O(g)\), the mole - ratio of \(C_2H_6\) to \(CO_2\) is \(2:4 = 1:2\).
If \(n_{C_2H_6}=7.97\ mol\), then \(n_{CO_2}=2\times n_{C_2H_6}\) (since for every 1 mole of \(C_2H_6\) burned, 2 moles of \(CO_2\) are produced). So \(n_{CO_2}=2\times7.97\ mol = 15.94\ mol\)

Step3: Use the ideal gas law at STP (\(P = 1\ atm\), \(T=273\ K\), \(R = 0.0821\ L\cdot atm/(mol\cdot K)\))

The ideal gas law is \(PV=nRT\). At STP, \(V=\frac{nRT}{P}\), and since \(P = 1\ atm\) and \(R = 0.0821\ L\cdot atm/(mol\cdot K)\), \(T = 273\ K\), for \(n\) moles of a gas \(V=n\times22.4\ L/mol\) (derived from \(V=\frac{n\times0.0821\ L\cdot atm/(mol\cdot K)\times273\ K}{1\ atm}\)).
For \(n_{CO_2}=15.94\ mol\), \(V_{CO_2}=15.94\ mol\times22.4\ L/mol\approx357\ L\)

Answer:

E. \(358\ L\) (the small difference in the final digit is due to rounding differences in the intermediate steps)