A. What is the total volume of gases produced at 819 K and 1.00 atm pressure whe

Question

A. What is the total volume of gases produced at 819 K and 1.00 atm pressure when 320 g of ammonium nitrite undergoes the following decomposition reaction? NH4NO2(s) N2(g) + 2H2O(g) B. What volume of ammonia gas, measured at 660.3 mmHg and 58.2oC, is required to produce 6.46 g of ammonium sulfate according to the following balanced chemical equation? 2NH3(g) + H2SO4(aq) (NH4)2SO4(s) C. If 379.8 mL of nitrogen gas, measured at 628.4 mmHg and 29.7oC, reacts with excess iodine according to the following reaction, what mass of nitrogen triiodide is produced? N2(g) + 3I2(s) 2NI3(s) Thanks

Explanation / Answer

A)

moles of NH4NO2 = 320 / 64.044 = 4.9966 mol

NH4NO2(s) -----------------> N2(g) + 2H2O(g)

moles of N2 = 4.9966

moles of H2O = 2 x 4.9966 = 9.993

total moles of gases = 4.9966 + 9.993 = 14.9897 mol

P V = n R T

1 x V = 14.9897 x 0.0821 x 819

V = 1007.9

total volume of gases = 1008 L

B)

2NH3(g) + H2SO4(aq) -------------> (NH4)2SO4(s)

moles of (NH4)2SO4 = 6.46 / 132.1395 = 0.04889

2 mol NH3 -------------- 1 mol (NH4)2SO4

??             -------------- 0.04889 mol (NH4)2SO4

moles of NH3 = 2 x 0.04889 = 0.097775

pressure = 660.3 / 760 = 0.8688

P V = n R T

0.8688 x V = 0.097775 x 0.0821 x 3312.

V = 3.06 L

volume of ammonia gas = 3.06 L

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