Question

A 0.4230 g sample of impure sodium nitrate (contains sodium nitrate plus inert ingredients) was heated, converting all the sodium nitrate to 0.2820 g of sodium nitrite and oxygen gas. Determine the percent of sodium nitrate in the original sample.

Answer 1

do you have the answer? hope this is correct.

Answer 2

Percent of sodium nitrate is 82.13%

Step-by-step explanation:

`NaNO_(3)\rightarrow NaNO_(2)+(1)/(2)O_(2)`

Number of moles of a compound is the ratio of mass of the compound to molar mass of the compound.

Molar mass of
`NaNO_(2)` = 68.9953 g/mol

Molar mass of
`NaNO_(3)` = 84.9947 g/mol

So, 0.2820 g
`NaNO_(2)` =
`(0.2820)/(68.9953)` moles of
`NaNO_(2)` = 0.004087 moles of
`NaNO_(2)`

Above equation suggests, 1 mol of
`NaNO_(2)` is formed from 1 mol of
`NaNO_(3)`

So, 0.004087 moles of
`NaNO_(2)` are formed from 0.004087 moles of
`NaNO_(3)`

So, mass of 0.004087 moles of
`NaNO_(3)` =
`(0.004087* 84.9947)g=0.3474g`
`NaNO_(3)`

% of
`NaNO_(3)` in sample =
`(0.3474)/(0.4230)* 100`% =
`82.13`%