# In 58*g of "ammonium nitrate", what mass of nitrogen is present?

May 31, 2017

Approx. $20 \cdot g$.

#### Explanation:

We work out a molar quantity of $\text{ammonium nitrate:}$

$\text{Moles of ammonia nitrate} = \frac{58.0 \cdot g}{80.05 \cdot g \cdot m o {l}^{-} 1} = 0.725 \cdot m o l$.

And we take this molar quantity, and multiply it by the appropriately modified molar mass of nitrogen...........

0.725*molxx2xx14.01*g*mol^-1=??*g