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Chemistry LibreTexts

Answers to Stoichiometry Problems

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  • 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. \(\mathrm{7.00\: g\: N_2 \times \left(\dfrac{1\: mol\: N_2}{28.0\: g\: N_2}\right) \times \left(\dfrac{6\: mol\: Li}{1\: mol\: N_2}\right) = 1.50\: mol\: Li}\)


    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. \(\mathrm{1.20 \times 10^{21}\: molecules\: H_2O \times \left(\dfrac{1\: mol\: H_2O}{6.02 \times 10^{23}\: molecules}\right) \times \left(\dfrac{1\: mol\: SnO_2}{2\: mol\: H_2O}\right) \times\left(\dfrac{150.7\: g\: SnO_2}{1\: mol\: SnO_2}\right)}\)
      \(\mathrm{= 0.150\: g\: SnO_2}\)
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