Question

how many moles of hydrochloric acid (in mol) are needed to produce 0.83 mol of iron(ii) chloride? fes + 2 hcl → fecl₂ + h₂s

Explanation:

Step1: Identify mole ratio

From the reaction $\ce{FeS + 2HCl -> FeCl_{2} + H_{2}S}$, the mole ratio of $\ce{HCl}$ to $\ce{FeCl_{2}}$ is $2:1$.

Step2: Calculate moles of HCl

Let $n(\ce{HCl})$ be moles of HCl, $n(\ce{FeCl_{2}}) = 0.83\space mol$.
Using the ratio: $\frac{n(\ce{HCl})}{n(\ce{FeCl_{2}})}=\frac{2}{1}$
So, $n(\ce{HCl}) = 2\times n(\ce{FeCl_{2}}) = 2\times0.83 = 1.66\space mol$.

Answer:

$1.66$