Question

10g of Fe powder was mixed with a $CuS{O}_4$ solution which contained 70g of $CuS{O}_4$. What is the weight of the copper obtained?

[Molar mass of $Fe=55.85g$ & Molar mass of $Cu=63.6g$]

• A. $11.39$
• B. $10.39$
• C. $13.39$
• D. $12.39$

Answer 1

The correct option is A

$11.39$

The question talks about reaction between $Fe$ and $CuS{O}_4$. Observe that weights of the reactants are given and the weight of a product is asked. Thus, it is a weight-weight stoichiometry problem.
In all such stoichiometry problems, start by balancing the reaction
$Fe$ + $CuS{O}_4$ $\to$ $FeS{O}_4$ + $Cu$
From the equation, 1 mole of $Fe$ = 1 mole of $Cu$
$\therefore 55.85g$ of $Fe$ $\approx 63.6g$ of $Cu$
$\therefore 10g$ of $Fe$ will displace 11.39g of Cu if available.
The solution has $70g$ of $CuS{O}_4$
1 mole $CuS{O}_4$ has $63.6g$ of Cu
i.e., $[63.6+32+16(4\left)\right]g$ of $CuS{O}_4$ $=63.6g$ of $Cu$

$159.6g$ of $CuS{O}_4$ $=63.6g$ of $Cu$
$70g$ of $CuS{O}_4=27.76g$ of $Cu$

$\therefore$ $10g$ of $Fe$ $=10\times \frac{63.6}{55.85}$ $=11.39g$ of $Cu$

$\therefore$ In the $CuS{O}_4$ solution, $27.76g$ of $Cu$ is present of which $11.39g$ of $Cu$ will be precipitated out by the addition of $10g$ of $Fe$.