Question

If 10.0 g NH403 (s) decomposes according to the reaction equation shown here, what volume of oxygen gas would be formed if it were measured at 78 degrees celsius and 740 torr? The reaction equation is:

4NH4O3--> 2NH4NO3 + O2 + 4H2O

Answer 1

4NH₄O₃--> 2NH₄NO₃ + O₂ + 4H₂O

Given, mass of NH₄O₃= 10 g

Molar mass of NH₄O₃= 66.04 g/mol

Now, convert the given mass of NH₄O₃ into number of moles-

Number of moles of NH₄O₃= mass of NH₄O₃/ molar mass of NH₄O₃ = 10 g / 66.04 g/mol = 0.151 mol

Therefore, total number of moles of NH₄O₃ reacted in the reaction are 0.151 mol.

According to reaction, 4 mol of NH₄O₃ gives 1 mol of O₂.

So, Number of moles of O₂ produced when 0.151 mol of NH₄O₃ decomposes is calculated as follows-

Moles of O₂ produced = (1.0 * 0.151) / 4 = 0.0378 mol

Total O₂ produced by decomposition of 10 g NH₄O₃ is 0.0378 mol.

According to ideal gas law, PV = nRT

P = 740 torr = 98658.6 Pa,

T = 78 ^oC = 351.15 K,

R = 8.314*10⁶ cm³.Pa K^-1 Mol^-1,

n= 0.0378 mol.

Put it in ideal gas equation to calculate V. Then we get,

V = nRT/P = (0.0378 mol . 8.314*10⁶ cm³.Pa K^-1 Mol^-1 . 351.15 K) / 98658.6 Pa = 1118.56 cm³

Therefore, decomposition of 10 g of NH₄O₃ produces 1118.58 cm3 of O₂ when measured at 78 degrees celsius and 740 torr.