Certainly! To solve this problem, let’s break down the steps systematically:
### Step 1: Convert the Volume of CO2 from mL to L
Given:
- Volume of CO2 = 23.6 mL
To convert milliliters to liters, we use the conversion factor:
[tex]\[ 1 \text{ mL} = 0.001 \text{ L} \][/tex]
So,
[tex]\[ \text{Volume of CO2 in liters} = 23.6 \text{ mL} \times 0.001 \text{ L/mL} = 0.0236 \text{ L} \][/tex]
### Step 2: Calculate Moles of CO2
At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 liters. Given the volume, we can calculate the number of moles of CO2:
[tex]\[ \text{Moles of CO2} = \frac{\text{Volume of CO2}}{\text{Molar Volume at STP}} = \frac{0.0236 \text{ L}}{22.4 \text{ L/mol}} \approx 0.001053571 \text{ mol} \][/tex]
### Step 3: Use Stoichiometry to Find Moles of NaHCO3
From the chemical equation:
[tex]\[ 2 \text{NaHCO}_3 (s) \xrightarrow{\Delta} \text{Na}_2\text{CO}_3 (s) + \text{H}_2\text{O} (l) + \text{CO}_2 (g) \][/tex]
2 moles of NaHCO3 produce 1 mole of CO2.
With the moles of CO2 calculated, we can find the moles of NaHCO3:
[tex]\[ \text{Moles of NaHCO3} = 2 \times \text{Moles of CO2} = 2 \times 0.001053571 \text{ mol} \approx 0.002107143 \text{ mol} \][/tex]
### Step 4: Calculate the Mass of NaHCO3
Given the molar mass of NaHCO3 is 84.01 g/mol, we can find the mass:
[tex]\[ \text{Mass of NaHCO3} = \text{Moles of NaHCO3} \times \text{Molar Mass of NaHCO3} = 0.002107143 \text{ mol} \times 84.01 \text{ g/mol} \approx 0.177021 \text{ g} \][/tex]
### Final Answer
The mass of sodium hydrogen carbonate that decomposes to give 23.6 mL of carbon dioxide gas at STP is:
[tex]\[ m_{\text{NaHCO3}} \approx 0.177 \text{ g} \][/tex]
So to summarize,
[tex]\[ m_{\text{NaHCO3}} = 0.177 \text{ g} \][/tex]
