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Given the equation:

[tex] 10 \text{Na} + 2 \text{NaNO}_3 \rightarrow 6 \text{Na}_2\text{O} + \text{N}_2 [/tex]

403 g of solid sodium react completely with an excess of sodium nitrate. How much of each substance was involved in the reaction?

**Column A:**

1. Yield of nitrogen gas in liters
2. Grams of sodium nitrate needed to complete the reaction
3. Grams of [tex]\text{Na}_2\text{O}[/tex] formed
4. Moles of sodium consumed

**Column B:**

a. 651
b. 17.5
c. 298
d. 39.3
e. 24.8
f. 793

Answer :

Sure! Let's go through the problem step-by-step.

We have the reaction:
[tex]\[ 10 \, \text{Na} + 2 \, \text{NaNO}_3 \rightarrow 6 \, \text{Na}_2\text{O} + \text{N}_2 \][/tex]

We start by noting that we are provided with 403 grams of sodium (Na) and need to find how much of each substance is involved when it reacts completely with excess sodium nitrate ([tex]\(\text{NaNO}_3\)[/tex]).

### Step 1: Calculate Moles of Sodium (Na) Consumed
Given the mass of sodium is 403 grams, we first calculate the moles of sodium using its molar mass, which is 22.99 g/mol.

[tex]\[
\text{Moles of Na} = \frac{403 \, \text{g}}{22.99 \, \text{g/mol}} \approx 17.5 \, \text{moles}
\][/tex]

### Step 2: Calculate Grams of Sodium Nitrate ([tex]\(\text{NaNO}_3\)[/tex]) Needed
From the balanced equation, 10 moles of sodium react with 2 moles of sodium nitrate. So, for every 10 moles of Na, we need 2 moles of NaNO[tex]\(_3\)[/tex].

Calculate the moles of NaNO[tex]\(_3\)[/tex] needed:

[tex]\[
\text{Moles of NaNO}_3 = \left(\frac{2}{10}\right) \times 17.5 \approx 3.5 \, \text{moles}
\][/tex]

Using the molar mass of NaNO[tex]\(_3\)[/tex] (85.00 g/mol), we find the grams needed:

[tex]\[
\text{Grams of NaNO}_3 = 3.5 \, \text{moles} \times 85.00 \, \text{g/mol} \approx 298 \, \text{grams}
\][/tex]

### Step 3: Calculate Grams of Sodium Oxide ([tex]\(\text{Na}_2\text{O}\)[/tex]) Formed
From the reaction, 10 moles of sodium produce 6 moles of sodium oxide. So, for every 10 moles of Na, we get 6 moles of Na[tex]\(_2\)[/tex]O.

Calculate the moles of Na[tex]\(_2\)[/tex]O formed:

[tex]\[
\text{Moles of Na}_2\text{O} = \left(\frac{6}{10}\right) \times 17.5 \approx 10.5 \, \text{moles}
\][/tex]

Using the molar mass of Na[tex]\(_2\)[/tex]O (61.98 g/mol), we find the grams formed:

[tex]\[
\text{Grams of Na}_2\text{O} = 10.5 \, \text{moles} \times 61.98 \, \text{g/mol} \approx 651 \, \text{grams}
\][/tex]

### Step 4: Calculate the Volume of Nitrogen Gas ([tex]\(\text{N}_2\)[/tex]) Produced
From the balanced equation, 10 moles of sodium yield 1 mole of nitrogen gas.

Calculate the moles of N[tex]\(_2\)[/tex] formed:

[tex]\[
\text{Moles of N}_2 = \frac{17.5}{10} \approx 1.75 \, \text{moles}
\][/tex]

Assuming standard temperature and pressure (STP) where 1 mole of gas occupies 22.414 liters, we find the volume:

[tex]\[
\text{Volume of N}_2 = 1.75 \, \text{moles} \times 22.414 \, \text{L/mol} \approx 39.3 \, \text{liters}
\][/tex]

### Final Answers
Putting it all together:

1. Yield of nitrogen gas in liters: ~39.3 liters
2. Grams of sodium nitrate needed: ~298 grams
3. Grams of [tex]\(\text{Na}_2\text{O}\)[/tex] formed: ~651 grams
4. Moles of sodium consumed: ~17.5 moles

These are the amounts of each substance involved in the reaction.

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