In the reaction shown, what is the mole ratio that would be used to determine the number of moles of H2 that would be produced when 3.5 moles of AIC13 are produced? 2Al(s) + 6HCl(aq) + 3H2 2AlCl3(aq) O 2 mol Al divided by 3 mol HCI O 3 mol H2 divided by 2 mol AICI: 3.5 mol AlCl3 divided by 3 mol of H2 O 3 mol AlCl3 divided by 2 mol H2 O 2 mol AlCl3 divided by 3 mol of H2​

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%