# How many moles of solid aluminum are needed to react with nitric acid in order to produce 5.0 moles of hydrogen gas?

Aug 2, 2017

A bit over $3 \cdot m o l$ aluminum are required...........

#### Explanation:

We need a stoichiometric equation that represents the oxidation of aluminum metal.....

$A l \left(s\right) + 3 H N {O}_{3} \left(a q\right) \rightarrow A l {\left(N {O}_{3}\right)}_{3} \left(a q\right) + \frac{3}{2} {H}_{2} \left(g\right) \uparrow$

And thus each mole of metal generates $\frac{3}{2}$ moles dihydrogen gas......And so if we want a $5 \cdot m o l$ quantity of ${H}_{2}$ we take the quotient......

$\frac{5 \cdot m o l}{\frac{3}{2} \cdot m o l} = \frac{10}{3} \cdot m o l$ $\text{aluminum metal}$. Given the molar mass of aluminum is $27.0 \cdot g$, what quantity of metal are needed for the given reaction?