Calculating Concentration

Question

Given the reaction: 2Al(s) + 3H2SO4(aq) ---> 3H2 + Al2(SO4)3(aq)

If aluminum is placed in a 450 mL solution of 1.35 M H2SO4. What mass of aluminum would be needed to completely react?

Answer 1

The key here is finding the moles of H2SO4, and using that to find the moles of Al by using the equation given in the question.

We start off with our 1.35 M solution of H2SO4. We have 450 mL of it, which is equal to .45 L. M is equal to moles/L, so we can multiply 1.35 mol/L x .45 L, the L cancel out, and we get .608 moles of H2SO4.

We now need to find the moles of Al needed to react, so looking at the equation given in the problem, we can see that there are 2 moles of Al for every 3 moles of H2SO4, so we take the .608 moles of H2SO4 and multiple it by 2 mol Al/3 mol H2SO4 and get .405 moles of Al.

We can find the molar mass of Al from a periodic table, which is about 26.98 g/mol. We take the .405 moles of Al and multiply it by 26.98 g/mol, and get 10.9 grams of Al!