Question

what quantity in moles of k₂s will be required to completely react with 77.0 ml of 0.200 m fecl₃? 3k₂s(aq) + 2fecl₃(aq) → fe₂s₃(s) + 6kcl(aq)

Explanation:

Step1: Calculate moles of FeCl₃

Use the formula $n = M\times V$. Given $V = 77.0\ mL=0.0770\ L$ and $M = 0.200\ M$.
$n_{FeCl_3}=0.200\ mol/L\times0.0770\ L = 0.0154\ mol$

Step2: Determine mole - ratio

From the balanced chemical equation $3K_2S(aq)+2FeCl_3(aq)\to Fe_2S_3(s) + 6KCl(aq)$, the mole - ratio of $K_2S$ to $FeCl_3$ is $\frac{n_{K_2S}}{n_{FeCl_3}}=\frac{3}{2}$.

Step3: Calculate moles of K₂S

$n_{K_2S}=\frac{3}{2}\times n_{FeCl_3}$. Substitute $n_{FeCl_3}=0.0154\ mol$ into the equation.
$n_{K_2S}=\frac{3}{2}\times0.0154\ mol = 0.0231\ mol$

Answer:

$0.0231\ mol$