Question

Information given: When heated to 350 ∘C at 0.950 atm, ammonium nitrate decomposes to produce nitrogen, water, and oxygen gases:
2NH4NO3(s)→2N2(g)+4H2O(g)+O2(g)

Question: How many liters of water vapor are produced when 23.6 g of NH4NO3 decomposes?
Express your answer with the appropriate units.
V=???

Thought it was....V =
13.216L but it's wrong

Answer 1

Answer on Question #56154 - Chemistry - General Chemistry

Question:

When heated to $350^{\circ}\mathrm{C}$ at 0.950 atm, ammonium nitrate decomposes to produce nitrogen, water, and oxygen gases:

$$2 \mathrm{NH_4NO_3}(s) \rightarrow 2 \mathrm{N_2}(g) + 4 \mathrm{H_2O}(g) + \mathrm{O_2}(g)$$

Question: How many liters of water vapor are produced when $23.6\mathrm{g}$ of $\mathrm{NH_4NO_3}$ decomposes?

Express your answer with the appropriate units.

Solution:

$$2 \mathrm{NH_4NO_3}(s) \rightarrow 2 \mathrm{N_2}(g) + 4 \mathrm{H_2O}(g) + \mathrm{O_2}(g)$$ $$n_{\mathrm{NH_4NO_3}} = \frac{m(\mathrm{NH_4NO_3})}{M_{\mathrm{NH_4NO_3}}} = \frac{23,6}{(18 + 14 + 48)\mathrm{g/mol}} = 0.295 \text{ moles}$$ $$n_{\mathrm{H_2O}} = 2 n_{\mathrm{NH_4NO_3}} = 0,59 \text{ moles}$$

According to the Mendeleev-Clapeyron law, value volume evolved during the reaction under the given conditions is:

$$V_{o_2} = \frac{n^* R^* T}{P} = \frac{0,59 \text{ mol}^* 0,082 \frac{\text{atm}^* \text{L}}{\text{mol}^* \text{K}} \cdot (350 + 273)\text{K}}{0,950 \text{ atm}} = 31,727 \text{ L}$$

where R - the universal gas constant; $a \cdot \frac{\text{atm}^* \text{L}}{\text{mol}^* \text{K}}$; T - temperature in Kelvins; P - pressure atm.

Answer: 31,727 L O₂

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