Question

Q2. (20 pts) 37.8 g of Y metal is allowed to react with 415 mL of an aqueous solution of HCl (d=1.088 g/mL) that contains 18.0% HCl by mass. Y() + HCl(aq) → YCl3(aq) + H2(g) (balance the reaction) a. What mass of H2(g) is produced? b. What mass of excess reactant remained at the end of this reaction? (Atomic weight of Y=26.98 g/mol, H=1.01g/mol, CI= 35.45 g/mol)

Answer 1

number of mols of Y metal = 37.8 g/26.98 g/mol = 1.401 mols

mass of HCl = 18 g
mols of HCl present = 18g/36.46 g/mol = 0.49369 mols

balanced reaction is
2Y + 6HCl --> 2YCl3 + 3H2(g)

since mols of HCl present is fewer, limiting reactant = HCl

mols of H2 gas formed = 3 mol H2/6mol HCl * 0.49369 mols HCl

mass of H2 formed = 3 mol H2/6mol HCl * 0.49369 mols HCl * 2.02 g/mol
= 0.4986 g

mols of Y metal reacted = 2 mol Y metal/6 mol HCl* 0.49369 mols HCl
= 0.16456 mols
mols of Y metal unreacted =1.401 mol - 0.16456 mols = 1.2364 mols

mass of Y metal unreacted = 1.2364 mols * 26.98 g/mol
= 33.35 g

answer
a) mass of H2 = 0.499 g
b) mass of metal = 33.4 g
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