# How many grams of solid zinc metal will completely react with 100.0 mL of 0.0525 M HCl, if zinc chloride and hydrogen gas are the only products?

Apr 29, 2017

Approx. $200 \cdot m g$.............

#### Explanation:

The first step is to write the stoichiometric equation, in order to establish the molar equivalence.........

$Z n \left(s\right) + 2 H C l \left(a q\right) \rightarrow Z n C {l}_{2} \left(a q\right) + {H}_{2} \left(g\right) \uparrow$

And then we use the relationship,

$\text{Concentration"="Moles of solute"/"Volume of solution} .$

And thus $\text{Moles of solute"="concentration"xx"volume of solution}$.

And so here..............

$\text{Moles of HCl} = 100 \cdot \cancel{m L} \times {10}^{-} 3 \cancel{L \cdot m {L}^{-} 1} \times 0.0525 \cdot m o l \cdot \cancel{{L}^{-} 1}$

$= 5.25 \times {10}^{-} 3 \cdot m o l .$

And thus we require a half equiv of zinc metal, i.e.

$= 5.25 \times {10}^{-} 3 \cdot m o l \times \frac{1}{2} \times 65.4 \cdot g \cdot m o {l}^{-} 1 = 0.171 \cdot g$ zinc metal........

But typically we would add a known stoichiometric EXCESS of hydrochloric acid, and either collect the dihydrogen gas evolved, OR back titrate with standardized $N a O H \left(a q\right)$ solution.