# Oxidation of zinc produced a 47.5*mL volume of dihydrogen at 1*atm and 273.15*K. What is the molar quantity of zinc that reacted?

Jun 14, 2017

I will do this for $1 \cdot a t m$, and $273.15 \cdot K$.........

#### Explanation:

We interrogate the redox reaction...........

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

We use the old Ideal gas equation to access the molar quantity....

$n = \frac{P V}{R T} = \frac{1 \cdot a t m \times 47.5 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1}{0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1 \times 273.15 \cdot K}$

$= 2.12 \times {10}^{-} 3 \cdot m o l$.

I am reluctant to use $\text{SATP}$ because the values seem to change across curricula. You will have to adapt this question to whatever values of $\text{SATP}$ apply. As a chemist, I tend to like measurement of pressure in $\text{atmospheres}$, and measurement of volume in $\text{millilitres}$. As a student you have to be adaptable.