Question

When ammonia burns in pure oxygen, the reaction is: 4 NH3(g) + 3 O2(g) 2 N2(g) + 6 H2O(g)What mass of water could be produced from 45.0 g of ammonia and excess Oxygen?

Answer 1

This is a stoichiometry problem, where we have an initial amount of reactant and we need to find out how much of the product will we end up with, in order to do that we need to:

1. Set up the properly balanced equation, which the question already provided us

4 NH3 + 3 O2 -> 2 N2 + 6 H2O

2. See how many moles of reactant there are in the given amount of grams, we will do that by using its molar mass

NH3 molar mass = 17g/mol, and we have 45 grams of it

17g = 1 mol

45g = x moles

x = 2.65 moles of NH3

3. Check the molar ratio between the two compounds, this we can see from the balanced equation, the molar ratio between NH3 and H2O is 4:6, this means that for every 4 moles of NH3 in the reaction, we will have 6 moles of water as product, but we have 2.65 moles of NH3

4 NH3 = 6 H2O

2.65 NH3 = x H2O

x = 3.97 moles of H2O, or we can go ahead and round up to 4 moles of H2O

4. Calculate how many grams will be equal to the number of moles that we found out, using the molar mass of H2O

H2O molar mass = 18g/mol

18 g = 1 mol

x grams = 4 moles of H2O

x = 72 grams of H2O