What is the maximum mass of H_2 that can be prepared if the mass of CH_4 is 983

Question

What is the maximum mass of H_2 that can be prepared if the mass of CH_4 is 983 g, the mass of water is 2010 g and CH_4 is the limiting reactant? The mass of H_2, in this case, depends on the amount of CH_4 since it is a limiting reactant Let's calculate the number of moles of hydrogen Now calculate the mass of H_2 using its number of moles and mass The reaction of methane and water is one way to prepare hydrogen for use as a fuel What mass of the excess reactant remains when the reaction is completed if the reactant is

Explanation / Answer

c) CH4(g) + H2O(g) CO(g) + 3H2(g)

The molar mass of methane, CH4, is (12.01 (C)) + (4 × 1.008 (H)) = 16.042. The number of moles of methane is therefore:

moles of methane = mass/molar mass = 983/16.042 =61.27 mol

The molar mass of water, H2O, is (2 × 1.008 (H)) + (16.00 (H)) = 18.016. The number of moles of water is therefore:

moles of water = 2010 / 18.016 = 111.56 mol

As the reaction is a 1:1 reaction of methane and water, methane(CH4) is the limiting reagent.

As the reaction is a 1:1 reaction, (111.56 – 61.27) = 50.29 mol of H2O will be left unreacted. This corresponds to a mass of:

mass of water = moles of water × molar mass = 50.29 × 18.016 = 906.02 g =0.906 kg.

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.