Question

how many grams of chromium are needed to react with an excess of CuSO4 to produce 27 g Cu

Answer 1

Task:

how many grams of chromium are needed to react with an excess of CuSO4 to produce 27 g Cu

Solution:

The chemical equation for this reaction is:

$$3 \mathrm{CuSO_4} + 2 \mathrm{Cr} = \mathrm{Cr_2(SO_4)_3} + 3 \mathrm{Cu}$$

The amount of Cu produced in reaction is

$$n(\mathrm{mol}) = \frac{m(g)}{MW(g/mol)}$$ $$n(\mathrm{Cu}) = \frac{27}{63.5} = 0.425 \mathrm{mol}$$

According to the chemical equation

$$n(\mathrm{Cr}) = \frac{3}{2} \cdot n(\mathrm{Cu})$$ $$n(\mathrm{Cr}) = \frac{3}{2} \cdot 0.425 = 0.283 \mathrm{mol}$$

The mass of Cr is

$$m(g) = n(\mathrm{mol}) \cdot MW(g/mol)$$

The molar weight of Cr is equal to its atomic weight in the periodic table of elements (MW(Cr) = 52 g/mol)

$$m(\mathrm{Cr}) = \frac{0.283}{52} = 14.7 \mathrm{g}$$

Answer: $m(\mathrm{Cr}) = 14.7 \mathrm{g}$