# Question 8e523

May 1, 2017

The answer is C: $1.5$ moles of ("NH"_4)_3"PO"_4#

#### Explanation:

in order to find the greatest mass of nitrogen, we can first see which compound has the largest number of mol of $\text{N}$.

To do this, we will multiply the no. of moles of each compound with the number of $\text{N}$ atoms in the compound and multiply the result with Avogadro's constant (do note that it is not necessary to get the exact numerical value as we can obtain the answer even by leaving the Avogadro's constant as it is)

For option a

$0.5 \times 2 \times \left(6.02 \times {10}^{23}\right) = \left(6.02 \times {10}^{23}\right)$

For option b

$2 \times \left(6.02 \times {10}^{23}\right)$

For option c

$1.5 \times 3 \times \left(6.02 \times {10}^{23}\right) = 4.5 \times \left(6.02 \times {10}^{23}\right)$

For option d

$2 \times 2 \times \left(6.02 \times {10}^{23}\right) = 4 \times \left(6.02 \times {10}^{23}\right)$

Since option C has the largest no. of moles of $\text{N}$, it will have the largest mass of $\text{N}$, too.