Question

Be sure to answer all parts. Industrially, hydrogen gas can be prepared by combining propane gas (CzHg) with steam at about 400°C. The products are carbon monoxide gas (CO) and hydrogen gas (H). (a) Write a balanced equation for the reaction. C, H,(8+3H,0(g) → 300(9)+7H,(8) (b) How many kilograms of H, can be obtained from 2.15 x 108 kg of propane? x 10 kg H2 gas

Answer 1

Stoichiometry of balance reaction indicates the relationship between reactant and products of any chemical reaction.

a) The balanced chemical equation for the reaction of propane gas with steam is:

C3H8(g)+3H2O(g)---- > 3CO(g)+7 H2(g)

b)

given that amount of propane = 2.15*10^3 kg

2.15*10^6 g C3H8

Number of moles = amount in g / molar mass

=   2.15*10^6 g C3H8 /44.0 g/ mole

= 4.89*10^4 mol C3H8

Now calculate moles of H2:

4.89*10^4 mol C3H8 *7 mole H2/1 mole C3H8

= 3.423 *10^5 Mol H2

Amount in g= number of mol* molar mass

= 3.423 *10^5 Mol H2 *2.016 g/ mole

= 6.9*10^5 g H2

6.9*10^2 kg H2

1000 g= 1.0 kg

Stoichiometry of balance reaction indicates the relationship between reactant and products of any chemical reaction a) The balanced chemical equation for the reaction of propane gas with steam is: C3H8(g)+3H20(g)---> 3CO(g)+7 H2(g) b) given that amount of propane 2.15*10^3 kg 2.15*10 6 g C3H8 Number of moles amount in g/molar mass 2.1510 6 g C3H8 /44.0 g/ mole =4.91 10^4 mool C3H8 Now calculate moles of H2: 4.91*10^4 mol C3H8 *11 mole H2/1 mole C3H8 5.401 *10^5 Mol H2 Amount in g= number of mol* molar mass = 5.401 *10^5 Mol H2 *2.016 g/ mole =1.09*10^6 g H2 1.09*10 3 kg H2 1000 g 1.0 kg