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Chapter 3: Problem 156

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

Short Answer

Expert verified
The percentage of sodium nitrate in the original sample is approximately \(83.36\%\).

Step by step solution

01

Write a balanced chemical equation

First, we need to write a balanced chemical equation for the decomposition of sodium nitrate (NaNO3) into sodium nitrite (NaNO2) and oxygen gas (O2). 2NaNO3 ⟶ 2NaNO2 + O2
02

Determine the molar mass of sodium nitrate and sodium nitrite

Next, we need to determine the molar mass of sodium nitrate (NaNO3) and sodium nitrite (NaNO2). NaNO3: 1(22.99) + 1(14.01) + 3(16.00) = 85.00 g/mol NaNO2: 1(22.99) + 1(14.01) + 2(16.00) = 69.00 g/mol
03

Convert the mass of sodium nitrite produced to moles

Now, let's convert the mass of sodium nitrite (0.2864 g) produced into moles using the molar mass of sodium nitrite (69.00 g/mol). moles of NaNO2 = mass of NaNO2 / molar mass of NaNO2 moles of NaNO2 = 0.2864 g / 69.00 g/mol = 0.00415 mol
04

Calculate the mass of sodium nitrate used

Using the balanced chemical equation, we can see that 1 mole of NaNO2 is produced from 1 mole of NaNO3. Therefore, the moles of NaNO3 used are equal to the moles of NaNO2 produced. moles of NaNO3 = 0.00415 mol Next, convert the moles of sodium nitrate to mass using the molar mass of sodium nitrate (85.00 g/mol). mass of NaNO3 = moles of NaNO3 * molar mass of NaNO3 mass of NaNO3 = 0.00415 mol * 85.00 g/mol = 0.35275 g
05

Determine the percentage of sodium nitrate in the original sample

Finally, we will determine the percentage of sodium nitrate in the original impure sample (0.4230 g). percentage = (mass of NaNO3 / mass of impure sample) * 100 percentage = (0.35275 g / 0.4230 g) * 100 = 83.36% Therefore, the percentage of sodium nitrate in the original sample is approximately \(83.36\%\).

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Most popular questions from this chapter

One of relatively few reactions that takes place directly between two solids at room temperature is $$ \mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}(s)+\mathrm{NH}_{4} \mathrm{SCN}(s) \longrightarrow $$ $$ \mathrm{Ba}(\mathrm{SCN})_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NH}_{3}(g) $$ In this equation, the \(\cdot 8 \mathrm{H}_{2} \mathrm{O}\) in \(\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}\) indicates the presence of eight water molecules. This compound is called barium hydroxide octahydrate. a. Balance the equation. b. What mass of ammonium thiocyanate (NH_sCN) must be used if it is to react completely with 6.5 g barium hydroxide octahydrate?

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