Answer
a. $70.0\text{ grams $N$}$
b. $140. \text{ grams $N$}$
c. $140.\text{ grams $N$}$
Work Step by Step
a. $5.00\text{ mols}*\frac{1\text{ mol $N$}}{1 \text{ mol $NH_3$}}*\frac{14.0067\text{ grams $N$}}{1\text{ mol $N$}}=70.0\text{ grams $N$}$
b. $5.00\text{ mols}*\frac{2 \text{ mol $N$}}{1 \text{ mol $N_2H_4$}}*\frac{14.0067\text{ grams $N$}}{1\text{ mol $N$}}=140. \text{ grams $N$}$
c. $5.00\text{ mols}*\frac{2\text{ mol $N$}}{1 \text{ mol $(NH_4)_2Cr_2O_7$}}*\frac{14.0067\text{ grams $N$}}{1\text{ mol $N$}}=140.\text{ grams $N$}$
