PLEASE HELP HONORS CHEMISTRY

Question

0.4230 g sample of impure sodium nitrate (contains sodium nitrate plus inert ingredients) was heated, converting all the sodium nitrate to 0.1720 g of sodium nitrite and oxygen gas. Determine the percent of sodium nitrate in the original sample.

Answer 1

Let's look at the balanced equation for the reaction taking place:

2NaNO₃ ==> 2NaNO₂ + O₂ ... balanced equation

moles NaNO₂ produced = 0.1720 g NaNO₂ x 1 mol NaNO₂ / 69.00 = 0.002493 moles NaNO₂

moles NaNO₃ originally present = 0.002493 mols NaNO₂ x 2 mol NaNO₃ / 2 mol NaNO₂ = 0.002493 mols NaNO₃ originally present

mass NaNO₃ originally present = 0.002493 mols NaNO₃ x 85.00 g /mol = 0.2119 g NaNO₃

%NaNO₃ = 0.2119 g / 0.4230 g (x100%) = 50.10%