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...

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.