You use 86.3 g NO and 25.6 g H_2 in an experiment and produce 49.0 g of ammonia. How much excess reactant was used to make the 49.0 g of ammonia?

Jun 10, 2017

Approx. $5 \cdot \frac{1}{2} \cdot m o l$ dihydrogen gas.........

Explanation:

We need (i) a stoichiometric equation.....

$N O \left(g\right) + \frac{5}{2} {H}_{2} \left(g\right) \rightarrow N {H}_{3} \left(g\right) + {H}_{2} O \left(g\right)$

Which (I think) is balanced, as indeed it must be if we purport to represent chemical reality.

And then (ii) we work out equivalent quantities of product and reactant.

$\text{Moles of NO} = \frac{86.3 \cdot g}{30.01 \cdot g \cdot m o {l}^{-} 1} = 2.88 \cdot m o l$

$\text{Moles of dihydrogen} = \frac{25.6 \cdot g}{2.02 \cdot g \cdot m o {l}^{-} 1} = 12.67 \cdot m o l$

$\text{Moles of ammonia} = \frac{49.0 \cdot g}{17.03 \cdot g \cdot m o {l}^{-} 1} = 2.88 \cdot m o l$

So you have got 100% yield. Call the newspapers!

The excess dihydrogen is simply {12.67-7.20}*mol=?*mol where, by the given stoichiometry, $7.2 \cdot m o l$ of dihydrogen gas actually reacted.

Capisce?