Question

If an electric discharge produces 800 cm3 of ozone (O3), how many cm3 of oxygen (O2) are required?

3O2(g) ---> 2O3(g)

2. When 75.0 dm3 of O2 react with an excess of glucose (C6H12O2), according to the reaction below, what volume of carbon dioxide will be produced?

6O2(g) + C6H12O6(s) ---> 6H2O(g) + 6CO2(g)

3. If an excess of nitrogen gas reacts with 250 L of hydrogen gas, according to the reaction below, how many L of ammonia will be produced?

N2(g) + 3H2(g) ---> 2NH3(g)

4. How many cm3 of oxygen would be required to react completely with 432 cm3 of hydrogen gas according to the reaction below?

2H2(g) + O2(g) ---> 2H2O(g)

Answer 1

Answer on Question #50437, Chemistry, Other

**Task:**

1) If an electric discharge produces $800~\mathrm{cm}^3$ of ozone $(\mathrm{O}_3)$, how many $\mathrm{cm}^3$ of oxygen $(\mathrm{O}_2)$ are required?

$$3 \mathrm{O}_2(\mathrm{g}) \rightarrow 2 \mathrm{O}_3(\mathrm{g})$$

2) When $75.0~\mathrm{dm}^3$ of $\mathrm{O}_2$ react with an excess of glucose $(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_2)$, according to the reaction below, what volume of carbon dioxide will be produced?

$$6 \mathrm{O}_2(\mathrm{g}) + \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6(\mathrm{s}) \rightarrow 6\mathrm{H}_2\mathrm{O}(\mathrm{g}) + 6\mathrm{CO}_2(\mathrm{g})$$

3) If an excess of nitrogen gas reacts with $250~\mathrm{L}$ of hydrogen gas, according to the reaction below, how many L of ammonia will be produced?

$$\mathrm{N}_2(\mathrm{g}) + 3\mathrm{H}_2(\mathrm{g}) \rightarrow 2\mathrm{NH}_3(\mathrm{g})$$

4) How many $\mathrm{cm}^3$ of oxygen would be required to react completely with $432~\mathrm{cm}^3$ of hydrogen gas according to the reaction below?

$$2\mathrm{H}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightarrow 2\mathrm{H}_2\mathrm{O}(\mathrm{g})$$

**Answer:**

$$\nu = \frac{V}{22.4}$$

1)

$$\nu(\mathrm{O}_3) = \frac{0.8}{22.4} = 0.036~\mathrm{mol}$$ $$\nu(\mathrm{O}_2) = \frac{3}{2} \nu(\mathrm{O}_3) = \frac{3}{2} \cdot 0.036 = 0.054~\mathrm{mol}$$ $$V(\mathrm{O}_2) = 22.4 \cdot \nu = 22.4 \cdot 0.054 = 1.21~\mathrm{l} = 1210~\mathrm{cm}^3$$ $$\nu = \frac{V}{22.4}$$

2)

$$\nu(\mathrm{O}_2) = \frac{75}{22.4} = 3.35~\mathrm{mol}$$ $$\nu(\mathrm{O}_2) = \nu(\mathrm{CO}_2) = 3.35~\mathrm{mol}$$ $$V(\mathrm{CO}_2) = 22.4 \cdot \nu = 22.4 \cdot 3.35 = 75~\mathrm{l} = 75~\mathrm{dm}^3$$ $$\nu = \frac{V}{22.4}$$

3)

$$\nu(\mathrm{H}_2) = \frac{250}{22.4} = 11.16~\mathrm{mol}$$ $$\nu(\mathrm{NH}_3) = \frac{2}{3} \nu(\mathrm{H}_2) = \frac{2}{3} \cdot 11.16 = 7.44~\mathrm{mol}$$ $$V(\mathrm{NH}_3) = 22.4 \cdot \nu = 22.4 \cdot 7.44 = 166.7~\mathrm{l}$$ $$\nu = \frac{V}{22.4}$$

4)

$$\nu(\mathrm{H}_2) = \frac{0.432}{22.4} = 0.02~\mathrm{mol}$$ $$\nu(\mathrm{O}_2) = \frac{1}{2} \nu(\mathrm{H}_2) = \frac{1}{2} \cdot 0.02 = 0.01~\mathrm{mol}$$ $$V(\mathrm{O}_2) = 22.4 \cdot \nu = 22.4 \cdot 0.01 = 0.224~\mathrm{l} = 224~\mathrm{cm}^3$$

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