# Answers to Stoichiometry Problems

1.

1. $$\mathrm{0.536\: mol\: Li \times \left(\dfrac{1\: mol\: N_2}{6\: mol\: Li}\right) \times\left(\dfrac{28.0\: g\: N_2}{1\: mol\: N_2} \right) = 2.50\: g\: N_2}$$
2. $$\mathrm{46.4\: g\: Li_3N \times \left(\dfrac{1\: mol\: Li_3N}{34.8\: g\: Li_3N}\right) \times \left(\dfrac{6\: mol\: Li}{2\: mol\: Li_3N}\right) = 4.00\: mol\: Li}$$
3. $$\mathrm{3.65\: g\: Li \times \left(\dfrac{1\: mol\: Li}{6.94\: g\: Li}\right) \times\left(\dfrac{2\: mol\: Li_3N}{6\: mol\: Li}\right) \times \left(\dfrac{34.8\: g\: Li_3N}{1\: mol\: Li_3N}\right) = 6.10\: g\: Li_3N}$$
4. $$\mathrm{7.00\: g\: N_2 \times \left(\dfrac{1\: mol\: N_2}{28.0\: g\: N_2}\right) \times \left(\dfrac{6\: mol\: Li}{1\: mol\: N_2}\right) = 1.50\: mol\: Li}$$

2.

1. $$\mathrm{0.211\: moles\: H_2 \times \left(\dfrac{1\: mol\: Sn}{2\: mol\: H_2}\right) \times \left(\dfrac{118.7\: g\: Sn}{1\: mol\: Sn}\right)=12.5\: g\: Sn}$$
2. $$\mathrm{339\: g\: SnO_2 \times \left(\dfrac{1\: mol\: SnO_2}{150.7\: g\: SnO_2}\right) \times \left(\dfrac{2\: mole\: H_2O}{1\: mol\: SnO_2}\right) = 4.50\: mol\: H_2O}$$
3. $$\mathrm{39.4\: g\: Sn \times \left(\dfrac{1\: mol\: Sn}{118.7\: g\: Sn}\right) \times \left(\dfrac{1\: mol\: SnO_2}{1\: mol\: Sn}\right) \times \left(\dfrac{150.7\: g\: SnO_2}{1\: mol\:SnO_2}\right)= 50.0\: g\: SnO_2}$$
4. $$\mathrm{3.00\: g\: H_2 \times \left(\dfrac{1\: mol\: H_2}{2.02\: g\: H_2}\right) \times \left(\dfrac{1\: mol\: Sn}{2\: mol\: H_2}\right) \times \left(\dfrac{6.02 \times10^{23}\: atoms}{1\: mol\: Sn}\right)= 4.47 \times 10^{23}\: atoms}$$
1. $$\mathrm{1.20 \times 10^{21}\: molecules\: H_2O \times \left(\dfrac{1\: mol\: H_2O}{6.02 \times 10^{23}\: molecules}\right) \times \left(\dfrac{1\: mol\: SnO_2}{2\: mol\: H_2O}\right) \times\left(\dfrac{150.7\: g\: SnO_2}{1\: mol\: SnO_2}\right)}$$
$$\mathrm{= 0.150\: g\: SnO_2}$$