Question

The ratio of the molar amounts of $H_{2}S$ needed to precipitate the metal ions from $20mL$ each of $1MCd(N{O}_3{)}_2$ and $0.5MCuS{O}_4$ is:

$(CdN{O}_3{)}_2+H_{2}S\to CdS+2HN{O}_3)$,

$CuS{O}_4+H_{2}S\to CuS+H_{2}S{O}_4)$

• A. $1:2$
• B. $1:1$
• C. $2:1$
• D. Indefinite

Answer 1

The correct option is C $2:1$

Calculating the number of moles

$Cd(N{O}_3{)}_2+H_{2}S\to CdS+2HN{O}_3$

Given molarity of $Cd(N{O}_3{)}_2=1M$

Volume of $Cd(N{O}_3{)}_2=20mL=0.02L$

$\text{Molarity}=\frac{\text{Number of moles of solute}}{\text{Volume of the solution (in L)}}$

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 $Cd(N{O}_3{)}_2=$𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 $\times$𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝑖𝑛 $L$)

$=1\times 0.02$

$=0.02mol$

According to the reaction,

$1mol$ of $Cd(N{O}_3{)}_2$ reacts with $1mol$ 𝑜𝑓 $H_{2}S$.

So, $0.02mol$ of $Cd(N{O}_3{)}_2$ will react with $0.02mol$ 𝑜𝑓 $H_{2}S$.

Calculating the molar amount of $H_{2}S$ needed

$CuS{O}_4+H_{2}S\to CuS+H_{2}S{O}_4$

Given molarity of $CuS{O}_4=0.5M$

Volume of $CuS{O}_4=20mL=0.02L$

𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 $CuS{O}_4=$𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 $\times$𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝑖𝑛 $L$) $=0.5\times 0.02$ $=0.01mol$

According to the reaction, $1mol$ of $CuS{O}_4$ reacts with

$1mol$ 𝑜𝑓 $H_{2}S$.

So, $0.01mol$ 𝑜𝑓 $CuS{O}_4$ will react with $0.01mol$ 𝑜𝑓 $H_{2}S$.

Hence, the ratio of molar amounts of $H_{2}S$ needed will be for $Cd(N{O}_3{)}_2$ and $CuS{O}_4$ is:

$=0.02:0.01$

$=2:1$

So, option (B) is the correct answer.