How many grams of hydrogen gas are required to produce ammonia with a mass of 12.80 g?

Question

3H_2 (g) + N_2 (g) ==> 2NH_3 (g)

please note that the underscore denotes subscript
(ex. H_2O

Answer 1

Assuming that the reaction is forward only (in reality, this is a reversible reaction),

First, we find the relative molecular mass of ammonia:

14.0 + 3(1.0) = 17.0

Now, we need to find out how many moles of ammonia is in 12.80 g of ammonia:

No. of moles of ammonia = 12.80/17.0 = 0.75294 moles.

From the equation given, 3 moles of hydrogen gives 2 moles of ammonia. So, what no. of moles of hydrogen gives 0.75294 moles of ammonia?

No. of moles of H_2 required = 3/2 x 0.75924 = 1.12941mol

Now, we have to convert this no. of moles of hydrogen to grams. The relative molecular mass of H_2 is 2. Therefore:

Mass of hydrogen gas required = 2 x 1.12941 = 2.259 g (correct to 4 significant figures) (Ans)

Answer 2

the key word is "excess." we don't care how much HCl is added since it's not the limiting reagent. we know that 75.0 moles of Zn are added to the reaction, so therefore 75.0 moles of ZnCl2 will be formed (since they are in a 1 to 1 ratio). Since 2 Cl atoms are used in that molecule, and hydrogen gas uses 2 hydrogen atoms, we can just come to the conclusion that 75.0 moles of hydrogen gas was formed since the reagents are proportionally equal. 136.28 * 75 moles = 10221 g

Answer 3

Convert the 12.80g of ammonia to moles by dividing by the molecular mass.

Now multiply by the molar ratio of 3/2 to get moles of H2. Now multiply by th emolecular mass of H2 for your answer

Answer 4

12.8 g NH3 * mol/17g NH3 = .75 mol NH3

.75 mol NH3 * 3 mol H2/2 mol NH3 = 1.125 mol H2

1.125 mol H2 * 2g/mol H2 = 2.25 g H2