Question

question 11
1 pts
industrially, hydrogen gas can be made by combining propane gas (c3h8) with water vapor at around 400°c. this forms carbon monoxide gas and hydrogen gas according to the following reaction: c3h8(g) + 3 h2o(g) → 3 co(g) + 7 h2(g). how many kilograms of hydrogen can be obtained if one begins with 3.55×103 kg of propane?
o 564 kg h2
o 2.76×103 kg h2
o 1.14×103 kg h2
o 4.47×104 kg h2
question 12
1 pts

Explanation:

Step1: Calculate the molar mass of propane ($C_3H_8$)

The molar mass of $C$ is $12.01\ g/mol$ and of $H$ is $1.008\ g/mol$. For $C_3H_8$, $M = 3\times12.01+8\times1.008=44.1\ g/mol$. The mass of propane is $m_{C_3H_8}=3.55\times 10^{3}\ kg = 3.55\times 10^{6}\ g$. The number of moles of propane, $n_{C_3H_8}=\frac{m_{C_3H_8}}{M}=\frac{3.55\times 10^{6}\ g}{44.1\ g/mol}\approx8.05\times 10^{4}\ mol$.

Step2: Use the stoichiometry of the reaction

From the reaction $C_3H_8(g)+3H_2O(g)\to3CO(g)+7H_2(g)$, the mole ratio of $C_3H_8$ to $H_2$ is $1:7$. So the number of moles of $H_2$, $n_{H_2}=7\times n_{C_3H_8}=7\times8.05\times 10^{4}\ mol = 5.64\times 10^{5}\ mol$.

Step3: Calculate the mass of $H_2$

The molar mass of $H_2$ is $M_{H_2}=2.016\ g/mol$. The mass of $H_2$, $m_{H_2}=n_{H_2}\times M_{H_2}=5.64\times 10^{5}\ mol\times2.016\ g/mol\approx1.14\times 10^{6}\ g = 1.14\times 10^{3}\ kg$.

Answer:

$1.14\times 10^{3}\ kg\ H_2$ (the second option in the given choices)