Question

I need help with the set up for this problem

Answer 1

The balanced chemical equation for the reaction is:

3 Ba(NO3)2(aq) + Al2(SO4)3(aq) → 3 BaSO4(s) + 2 Al(NO3)3(aq)

According to the equation, 1 mole of Al2(SO4)3 reacts to form 2 moles of Al(NO3)3.

Given that 0.100 mol of Al2(SO4)3 reacts, we can determine the amount of Al(NO3)3 formed:

0.100 mol Al2(SO4)3 * (2 mol Al(NO3)3 / 1 mol Al2(SO4)3) = 0.200 mol Al(NO3)3

Now we can calculate the mass of aluminum nitrate formed using its molar mass:

0.200 mol Al(NO3)3 * 213.01 g/mol = 42.602 g

Therefore, approximately 42.602 grams of aluminum nitrate can form during the reaction.

I hope this helps! :)