In this stoichiometry question, we have the following reaction:
4 Al + 3 N2 -> 2 Al2N3
We have:
2.5 grams of N2
We want to know how much Al we need in order to react with 2.5 grams of N2, and in order to do that, we need to find out how many moles of N2 we have in 2.5 grams, we will be using its molar mass, 28g/mol to do it:
28g = 1 mol
2.5g = x moles
28x = 2.5
x = 2.5/28
x = 0.089 moles of N2 in 2.5 grams
Now according to the molar ratio, we have 4 moles of Al for every 3 moles of N2, therefore, if we have 0.089 moles of N2:
4 Al = 3 N2
x Al = 0.089 N2
x = 0.119 moles of Al
Now we can find the mass needed, using the number of moles, 0.119 moles, and also the molar mass, 27g/mol:
27g = 1 mol
x grams = 0.119 moles of Al
x = 3.2 grams of Al is required
Answer is 3.2 grams