With the reaction, determine the masses of $H S O_{3} N H_{2}$ $HSO_3NH_2$ required to produce 0.135 g of $N_{2} (g)$ $N_2(g)$ collected in a burette above water. How do you calculate the volume that this mass of $N_{2} (g)$ $N_2(g)$ would occupy at 23.4°C and $7.10 \cdot 10^{2}$ $7.10*10^2$ mmHg barometric pressure?

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With the reaction, determine the masses of $H S O_{3} N H_{2}$ $HSO_3NH_2$ required to produce 0.135 g of $N_{2} (g)$ $N_2(g)$ collected in a burette above water. How do you calculate the volume that this mass of $N_{2} (g)$ $N_2(g)$ would occupy at 23.4°C and $7.10 \cdot 10^{2}$ $7.10*10^2$ mmHg barometric pressure?

$H S O_{3} N H_{2} (a q) + N a N O_{2} (a q) \to N a H S O_{4} (a q) + N_{2} (g) + H_{2} O (l)$ $HSO_3NH_2(aq) + NaNO_2(aq) -> NaHSO_4(aq) + N_2(g) + H_2O(l)$

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Answer 1 · 2017-04-09T00:24:30

Calculate the masses from the related molar amounts. Calculate the volume of $N_{2}$ $N_2$ from the ideal gas laws.

Explanation:

The balanced equation shows that every mole of $N_{2}$ $N_2$ requires one mole of $H S O_{3} N H_{2}$ $HSO_3NH_2$ . 0.135g of $N_{2}$ $N_2$ is $(\frac{0.135}{28}) = 0.00482$ $(0.135/28) = 0.00482$ moles of $N_{2}$ $N_2$ .

This requires $(0.00482 \cdot 97) = 0.468 g$ $(0.00482 * 97) = 0.468g$ of $H S O_{3} N H_{2}$ $HSO_3NH_2$ and $(0.00482 \cdot 53) = 0.42555 g$ $(0.00482 * 53) = 0.42555g$ of $N a N O_{2}$ $NaNO_2$ .

The volume of $N_{2}$ $N_2$ produced is $V = (\frac{n \cdot R \cdot T}{P})$ $V = ((n*R*T)/P)$

P = 710/760 = 0.934atm, R = 0.0821 L-atm/K-mol, T = 296.6’K, n = 0.00482 moles, V = Liters

$V = (\frac{0.00482 \cdot 0.0821 \cdot 296.6}{0.934})$ $V = ((0.00482*0.0821*296.6)/0.934)$ ; $V = 0.126 L$ $V = 0.126L$

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$H S O_{3} N H_{2} (a q) + N a N O_{2} (a q) \to N a H S O_{4} (a q) + N_{2} (g) + H_{2} O (l)$ $HSO_3NH_2(aq) + NaNO_2(aq) -> NaHSO_4(aq) + N_2(g) + H_2O(l)$

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With the reaction, determine the masses of HSO3NH2HSO_3NH_2 required to produce 0.135 g of N2(g)N_2(g) collected in a burette above water. How do you calculate the volume that this mass of N2(g)N_2(g) would occupy at 23.4°C and 7.10⋅1027.10*10^2 mmHg barometric pressure?

HSO3NH2(aq)+NaNO2(aq)→NaHSO4(aq)+N2(g)+H2O(l)HSO_3NH_2(aq) + NaNO_2(aq) -> NaHSO_4(aq) + N_2(g) + H_2O(l)

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$H S O_{3} N H_{2} (a q) + N a N O_{2} (a q) \to N a H S O_{4} (a q) + N_{2} (g) + H_{2} O (l)$ $HSO_3NH_2(aq) + NaNO_2(aq) -> NaHSO_4(aq) + N_2(g) + H_2O(l)$