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

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

b)
`2.64* 10^3kg`

Step-by-step 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}=(8.31* 10^6g)/(44.1g/mol)=0.188* 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* 10^6moles` moles of
`C_3H_8` will give =
`(7)/(1)* 0.188* 10^6=1.32* 10^6moles` of
`H_2`

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

Thus
`2.64* 10^3kg` of
`H_2` can be obtained from
`8.31* 10^3` kg of propane

Answer 2

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.