Zn2s + 4 AgNO3ag - > 2 Zn (NO3) 2ag + 4 Ags

What amount (in moles) of silver nitrate will react with excess zinc to produce 100.0 g of silver metal?

Determine the mass of zinc nitrate that will be produced by 64.50 moles of zinc metal.

What mass of zinc metal is required to produce 45.00 g of silver metal?

- 0.927 moles of AgNO₃

- 87.7 g of Zn (NO₃) ₂

- 13.6 g of Zn

Explanation:

The reaction is:

2Zn (s) + 4AgNO₃ (aq) → 2Zn (NO₃) ₂ (aq) + 4 Ag (s)

First of all we need to determine the moles of silver that were produced

100 g / 107.87 g/mol = 0.927 moles of Ag

As the limiting reactant is the nitrate, we determine that ratio is 4:4

If I made 4 moles of silver, I used 4 moles of nitrate. In conclusion, 0.927 moles of Ag were made by 0.927 moles of nitrate

If we still assume, that the Zn is in excess we use the 0.927 moles of silver nitrate. Ratio is 4:2 so the rule of three will be:

4 moles of silver nitrate can produce 2 moles of zinc nitrate

0.927 moles of silver nitrate will produce (0.927. 2) / 4 = 0.463 moles of zinc nitrate. We convert the moles to mass → 0.463 mol. 189.41 g / 1mol = 87.7 g

We need to determine the moles of silver that are produced

45 g / 107.87 g/mol = 0.417 moles

As the zn is in excess, we used the silver nitrate as the limiting reagent. Ratio was 4:4, so 0.417 moles were produced by 0.417 moles of AgNO₃

By stoichiometry, 4 moles of silver nitrate react with 2 moles of solid zinc. The rule of three will be: 4 moles of silver nitrate react with 2 moles of Zn

Then, 0.417 moles of AgNO₃ will react with (0.417. 2) / 4 = 0.208 moles of Zn

We convert the moles to mass → 0.208 mol. 65.41 g / 1mol = 13.6 g