Question

what is the mass of silver that can be prepared from 1.00 g of copper metal ?

Answer 1

What is the mass of silver that can be prepared from $1.00\mathrm{g}$ of copper metal?

Solution:

Write a balanced chemical equation describing the reaction. Arrange the coefficients in the reaction equation.

$$2 \mathrm{AgNO_3} + \mathrm{Cu} = \mathrm{Cu(NO_3)_2} + \mathrm{Ag} \downarrow$$ $$2 \mathrm{Ag^+} + \mathrm{Cu} = \mathrm{Cu^{2+}} + 2 \mathrm{Ag}$$

We will find the amount of substance of copper, that may need for the receipt of silver.

$$n (Cu) = \frac{m (Cu)}{M (Cu)}$$ $$n (Cu) = \frac{1 \, g}{63,75 \, g / mol} = 0,0157 \, mol$$

Refer to the balanced equation. 2 moles of Silver can be made from 1 mole of Copper, so the ratio is 2 to 1. This means the 0,0157 moles of Copper will produce $2 \times 0,0157 = 0,0314$ mol of Silver:

$$\frac{n (Cu)}{n (Ag)} = \frac{1}{2}$$ $$n (Ag) = n (Cu) \times 2$$ $$n (Ag) = 2 \times 0,0157 = 0,0314 \, mol$$

The weight of Silver can be calculated from its molar mass:

$$m (Ag) = n (Ag) \times M (Ag)$$ $$m (Ag) = 0,0314 \, mol \times 107,87 \, g / mol = 3,39 \, g$$

**Answer:** 3,39 grams of Silver