Question

How many grams of copper will be replaced in $2L$ of a $1.50MCuS{O}_4$ solution if the latter is made to react with $27.0g$ of aluminium?
(Atomic mass of $Cu=63.5,$ Atomic mass of $Al=27.0$)

$(3CuS{O}_4+2Al\to A{l}_2(S{O}_4{)}_3+3Cu)$

• A. $10g$
• B. $95.25g$
• C. $190g$
• D. $48g$

Answer 1

The correct option is B $95.25g$

$\text{Calculating moles of}CuS{O}_4\text{and aluminum}$

Given $2L$ of $1.5MCuS{O}_4$ solution,

Thus, number of moles of $CuS{O}_4$

$=Molarity\times Volume\left(inL\right)$

$=3mol$

And $27g$ of aluminum is reacted

Thus no.of moles of $Al=\frac{mass}{molarmass}$

$=\frac{27g}{27gmo{l}^{-1}}=1mol$

$\text{Finding the amount of}CuS{O}_4\text{reacted with}Al$

Chemical reaction given:

$3CuS{O}_4+2Al\to A{l}_2(S{O}_4{)}_3+3Cu$

Thus, according to the reaction:

$2mol$ of $Al$ reacts with 3 mol of $CuS{O}_4$

1 mol of $Al$ will react with $\frac{3}{2}$ mol of $CuS{O}_4$

So, number of mole of $Cu$ replaced is also equal to $\frac{3}{2}mol.$

Thus, mass of copper replaced

$=no.ofmoles\times molarmass$

$=\frac{3}{2}\times 63.5g$

$=45.25g$

So, option (b) is the correct answer.