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Chapter 3: Problem 146

A \(0.4230-\mathrm{g}\) sample of impure sodium nitrate was heated, converting all the sodium nitrate to \(0.2864 \mathrm{~g}\) of sodium nitrite and oxygen gas. Determine the percent of sodium nitrate in the oriainal sample.

Short Answer

Expert verified
The percentage of sodium nitrate in the original sample is 83.4%.

Step by step solution

01

Write down the balanced chemical equation for the decomposition of sodium nitrate.

The decomposition of sodium nitrate (NaNO3) produces sodium nitrite (NaNO2) and oxygen gas (O2). The balanced chemical equation is: \[2 \mathrm{NaNO}_3 \rightarrow 2 \mathrm{NaNO}_2 + \mathrm{O}_2\]
02

Calculate the moles of sodium nitrite formed.

We are given that 0.2864 g of sodium nitrite was formed after heating. To find the moles of sodium nitrite, we need to use its molar mass, which is 69.0 g/mol. Moles of NaNO2 formed = mass of NaNO2 / molar mass of NaNO2 Moles of NaNO2 formed = \(0.2864 \mathrm{~g} / 69.0 \mathrm{~g/mol}\) Moles of NaNO2 formed = \(4.15 \times 10^{-3} \mathrm{~mol}\)
03

Use stoichiometry to find the moles of sodium nitrate in the original sample.

From the balanced chemical equation, we can see that 2 moles of sodium nitrate decompose to produce 2 moles of sodium nitrite. Therefore, the moles of sodium nitrate in the original sample should be equal to the moles of sodium nitrite formed. Moles of NaNO3 = Moles of NaNO2 = \(4.15 \times 10^{-3} \mathrm{~mol}\)
04

Convert the moles of sodium nitrate to grams.

To convert the moles of sodium nitrate to grams, we need its molar mass, which is 85.0 g/mol. Mass of NaNO3 = moles of NaNO3 × molar mass of NaNO3 Mass of NaNO3 = \(4.15 \times 10^{-3} \mathrm{~mol} \times 85.0 \mathrm{~g/mol}\) Mass of NaNO3 = 0.3528 g
05

Calculate the percentage of sodium nitrate in the original sample.

Now that we have the mass of sodium nitrate in the original sample, we can calculate its percentage by dividing it by the total mass of the impure sample and multiplying by 100. Percentage of NaNO3 = (mass of NaNO3 / mass of impure sample) × 100 Percentage of NaNO3 = \((0.3528 \mathrm{~g} / 0.4230 \mathrm{~g}) \times 100\) Percentage of NaNO3 = 83.4% The percentage of sodium nitrate in the original sample is 83.4%.

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