Answer
a. 70.1 g of nitrogen
b. 140. g of nitrogen
c. 140. g of nitrogen
Work Step by Step
a. - Each $ NH_3 $ has 1 N atoms, thus:$$ 5.00 \space mole \space NH_4 \times \frac{ 1 \space moles \ N }{1 \space mole \space NH_3 } = 5.00 \space moles \space N $$
$ N $ : 14.01 g/mol
- Using the molar mass as a conversion factor, find the mass in g:
$$ 5.00 \space mole \times \frac{ 14.01 \space g}{1 \space mole} = 70.1 \space g$$
b. - Each $ N_2H_4 $ has 2 N atoms, thus:$$ 5.00 \space mole \space N_2H_4 \times \frac{ 2 \space moles \ N }{1 \space mole \space N_2H_4 } = 10.0 \space moles \space N $$
$$ 10.0 \space mole \times \frac{ 14.01 \space g}{1 \space mole} = 140. \space g$$
c.- Each $ N_2H_8Cr_2O_7 $ has 2 N atoms, thus:$$ 5.00 \space mole \space N_2H_8Cr_2O_7 \times \frac{ 2 \space moles \ N }{1 \space mole \space N_2H_8Cr_2O_7 } = 10.0 \space moles \space N $$
$$ 10.0 \space mole \times \frac{ 14.01 \space g}{1 \space mole} = 140. \space g$$
