Question

Be sure to answer all parts. Industrially, hydrogen gas can be prepared by combining propane gas (c3h8) with steam at about 400°c. The products are carbon monoxide (co) and hydrogen gas (h2). (a) write a balanced equation for the reaction. Include phase abbreviations. (b) how many kilograms of h2 can be obtained from 8.31 × 103 kg of propane

Answer 1

Balanced chemical reaction:

C₃H₈(g) + 3H₂O(g) → 3CO(g) + 7H₂(g).

M(C₃H₈) = 44.1 g/mol; molar mass of propane.

M(H₂) = 2 g/mol; molar mass of hydrogen.

From balanced chemical reaction: n(C₃H₈) : n(H₂) = 1 : 7.

7m(C₃H₈) : M(C₃H₈) = m(H₂) : M(H₂).

7·8310 kg : 44.1 g/mol = m(H₂) : 2 g/mol.

m(H₂) = 2638.09 kg; mass of hydrogen.

Answer 2

Answer: a) $C_3H_8(g)+3H_2O(g)\rightarrow 3CO(g)+7H_2(g)$

b) $2.64\times 10^3kg$

Explanation:

a) According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

$C_3H_8(g)+3H_2O(g)\rightarrow 3CO(g)+7H_2(g)$

b) $\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Number of moles of propane}=\frac{8.31\times 10^6g}{44.1g/mol}=0.188\times 10^6moles$

$C_3H_8(g)+3H_2O(g)\rightarrow 3CO(g)+7H_2(g)$

According to stoichiometry:

1 mole of $C_3H_8$ gives 7 moles of $H_2$

Thus $0.188\times 10^6moles$  moles of $C_3H_8$ will give =$\frac{7}{1}\times 0.188\times 10^6=1.32\times 10^6moles$ of $H_2$

Mass of $H_2=moles\times {\text {molar Mass}}=1.32\times 10^6moles\times 2g/mol=2.64\times 10^6g=2.64\times 10^3kg$

Thus $2.64\times 10^3kg$ of $H_2$ can be obtained from $8.31\times 10^3$ kg of propane