2NaNO3 → 2NaNO2 + O2. From the equation above, what is the volume of oxygen produced at s.t.p. when 8.5g of 2NaNO3 is heated until no further gas is evolved...
2NaNO3 → 2NaNO2 + O2. From the equation above, what is the volume of oxygen produced at s.t.p. when 8.5g of 2NaNO3 is heated until no further gas is evolved? (2NaNO3 = 85, molar volume of gases at s.t.p. = 22.4dm3)
Answer Details
The balanced chemical equation is 2NaNO3 → 2NaNO2 + O2. It shows that 2 moles of NaNO3 produces 1 mole of O2. We are given 8.5g of 2NaNO3. To find the number of moles, we divide the mass by the molar mass. Molar mass of 2NaNO3 = (2 x 23) + 2(14 + 3x16) = 46 + 2(14 + 48) = 46 + 124 = 170g/mol. Number of moles of 2NaNO3 = 8.5g / 170g/mol = 0.05mol. Since 2 moles of NaNO3 produces 1 mole of O2, the number of moles of O2 produced is also 0.05mol. The molar volume of gases at s.t.p. is 22.4dm3/mol. Therefore, the volume of O2 produced at s.t.p. is: Volume of O2 = Number of moles x Molar volume of gases at s.t.p. = 0.05mol x 22.4dm3/mol = 1.12dm3 Therefore, the volume of oxygen produced at s.t.p. when 8.5g of 2NaNO3 is heated until no further gas is evolved is 1.12dm3. Hence, the correct option is 1.12dm3.