Answer to Question #170067 in General Chemistry for Lari

Question

Please with the steps..

Consider the mixing of 5.1 L of  CO2 (g) and 3.1 L of H2O (g) at 45°C and 850 mm Hg. Determine the mass of O2 (g) that can be produced from the unbalanced reaction:

CO2 (g) + H2O (g) = C4H10 (l) +O2 (g)

Mass= ? in grams

Answer 1

To determine the mass of oxygen, the number of moles of gaseous $CO_2$ and $H_2O$ have to be computed based on the ideal gas law:

$pV=nRT$

$n=\frac{pV}{RT}\\ n(CO_2)=\frac{850\times 133 \times 10^{-3} Pa \times 5.1 L}{8.314 J \cdot K^{-1} \cdot mol^{-1} * (45+273) K} = 0.22 mol \\ n(H_2O)=\frac{850\times 133 \times 10^{-3} Pa \times 3.1 L}{8.314 J \cdot K^{-1} \cdot mol^{-1} * (45+273) K} = 0.13 mol$

If we want to know the mass of oxygen that can be produced from the reaction, it has to be balanced first:

$8CO_2 (g) + 10H_2O (g) =2C_4H_{10} (l) +13O_2 (g)$

The limiting reagent is water:

$CO_2 : 0.22/8 = 0.027\\ H_2O: 0.13 / 10 = 0.013$

Now, the mass of oxygen is:

$m(O_2) = n (O_2)\times M (O_2)= \frac{13\times 0.13}{10} mol \times 32 = 5.41 g$