Question

Sodium Carbonate decomposes

If 57.288 Liters of Carbon Dioxide at STP are produced, how many grams of Sodium
Carbonate decomposed?

Answer 1

To determine the amount of sodium carbonate decomposed when 57.288 liters of carbon dioxide at STP are produced, you can follow these steps:

1. **Calculate the number of moles of CO2 produced**:

- At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters.

- Convert 57.288 liters of CO2 to moles using the ratio: 1 mole = 22.4 liters.

2. **Write the balanced chemical equation for the decomposition of sodium carbonate**:

- The decomposition of sodium carbonate produces carbon dioxide and sodium oxide.

3. **Determine the stoichiometry of the reaction**:

- From the balanced equation, identify the mole ratio between CO2 and Na2CO3.

4. **Calculate the number of moles of sodium carbonate decomposed**:

- Use the mole ratio from the balanced equation to find the moles of Na2CO3 decomposed.

5. **Convert moles of sodium carbonate to grams**:

- Determine the molar mass of Na2CO3 (sodium carbonate).

- Multiply the moles of Na2CO3 decomposed by its molar mass to find the grams decomposed.

By following these steps, you can calculate the amount of sodium carbonate decomposed in grams based on the volume of carbon dioxide produced at STP. This approach allows you to apply stoichiometry principles to determine the quantity of reactant decomposed in a chemical reaction.

Answer 2

Answer:

265 grams

Explanation:

To solve this problem, we need to use stoichiometry to relate the volume of carbon dioxide produced to the amount of sodium carbonate decomposed.

First, we need to write the balanced chemical equation for the decomposition of sodium carbonate ($Na_2CO_3$):

$Na_2CO_3 \longrightarrow Na_2O + CO_2$

From the balanced equation, we see that 1 mole of sodium carbonate produces 1 mole of carbon dioxide.

Now, let's use the ideal gas law to relate the volume of carbon dioxide produced to the number of moles of carbon dioxide:

$PV = nRT$

Since the volume of carbon dioxide is given at STP (Standard Temperature and Pressure), we know that 1 mole of any gas at STP occupies 22.414 liters.

Therefore, we can use this conversion factor to find the number of moles of carbon dioxide produced:

$n = \dfrac{PV}{RT}$

Given:

• $P = 1$ atm (STP)
• $V = 57.288$ liters
• $R = 0.0821$ L atm/mol K (universal gas constant)
• $T = 273.15$ K (STP)

Now, we can calculate the number of moles of carbon dioxide produced:

$n = \dfrac{(1 \, \textsf{atm}) \times (57.288 \, \textsf{liters})}{(0.0821 \, \textsf{L atm/mol K}) \times (273.15 \, \textsf{K})}$

$n \approx \dfrac{57.288}{22.425615}$

$n \approx 2.5$

Now that we know the number of moles of carbon dioxide produced, and since the stoichiometry of the reaction is 1:1 for sodium carbonate to carbon dioxide, we can say that 2.5 moles of sodium carbonate decomposed.

The molar mass of sodium carbonate ($Na_2CO_3$) is approximately 106 g/mol.

Now, we can calculate the mass of sodium carbonate decomposed:

$\textsf{Mass} = \textsf{Number of moles} \times \textsf{Molar mass}$

$\textsf{Mass} = 2.5 \, \textsf{moles} \times 106 \, \textsf{g/mol}$

$\textsf{Mass} \approx 265 \, \textsf{grams}$

Therefore, approximately 265:grams of sodium carbonate decomposed to produce 57.288 liters of carbon dioxide at STP.