Question

According to the reaction below, how many grams of zinc, Zn, could react in 147.96 mL of 1.778 M AgNO3?

Zn(s) + 2 AgNO3(aq) → 2 Ag(s) + Zn(NO3)2(aq)

Answer 1

Molarity = 1.778 M

Volume = 147.96 mL = 0.14796 L

Moles = molarity X volume in L

n = 1.778 X 0.14796

n = 0.2631 mol

From the balanced equation given,

2 mol of silver nitrate needs 1 mol of zinc

Then,

0.2631 mol of silver nitrate needs 0.2631 / 2 = 0.1315 mol of zinc.

Moles of zinc = 0.1315 mol

Molar mass of zinc = 65.38 g/mol

Mass of zinc = moles X molar mass

m = 0.1315 X 65.38

m = 8.600 g.

Therefore,

Mass of zinc required = 8.600 g.