for the volume of carbon dioxide produced from 2100 kg of sodium hydrogen carbonate. The reaction is given…

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for the volume of carbon dioxide produced from 2100 kg of sodium hydrogen carbonate. The reaction is given as:

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The problem asks for the volume of carbon dioxide produced from 2100 kg of sodium hydrogen carbonate. The reaction is given as:

2\text{NaHCO}_3(\text{s}) \xrightarrow{\text{heat}} \text{Na}_2\text{CO}_3(\text{s}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{g})

We are given the relative atomic masses: $\text{Na}=23$ , $\text{H}=1$ , $\text{C}=12$ , $\text{O}=16$ .
We need to find the volume of $\text{CO}_2$ at STP (Standard Temperature and Pressure). At STP, 1 mole of any gas occupies 22.4 $\text{dm}^3$ .

Calculate the molar mass of $\text{NaHCO}_3$ .
$\text{Molar mass of } \text{NaHCO}_3 = (23 \times 1) + (1 \times 1) + (12 \times 1) + (16 \times 3)$ $= 23 + 1 + 12 + 48 = 84 \text{ g/mol}$
Convert the mass of $\text{NaHCO}_3$ from kg to g and then to moles.
Given mass of $\text{NaHCO}_3 = 2100 \text{ kg} = 2100 \times 1000 \text{ g} = 2.1 \times 10^6 \text{ g}$
$\text{Moles of } \text{NaHCO}_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{2.1 \times 10^6 \text{ g}}{84 \text{ g/mol}}$ $= 25000 \text{ mol}$
Use the stoichiometry of the reaction to find the moles of $\text{CO}_2$ produced.
From the balanced equation, 2 moles of $\text{NaHCO}_3$ produce 1 mole of $\text{CO}_2$ .
$\text{Moles of } \text{CO}_2 = \text{Moles of } \text{NaHCO}_3 \times \frac{1 \text{ mol } \text{CO}_2}{2 \text{ mol } \text{NaHCO}_3}$ $= 25000 \text{ mol} \times \frac{1}{2} = 12500 \text{ mol}$
Calculate the volume of $\text{CO}_2$ at STP.
At STP, 1 mole of gas occupies 22.4 $\text{dm}^3$ .
$\text{Volume of } \text{CO}_2 = \text{Moles of } \text{CO}_2 \times 22.4 \text{ dm}^3/\text{mol}$ $= 12500 \text{ mol} \times 22.4 \text{ dm}^3/\text{mol}$ $= 280000 \text{ dm}^3$
Convert the volume to $\text{m}^3$ (optional, but often preferred for large volumes).
Since $1 \text{ m}^3 = 1000 \text{ dm}^3$ :
$\text{Volume of } \text{CO}_2 = \frac{280000 \text{ dm}^3}{1000 \text{ dm}^3/\text{m}^3} = 280 \text{ m}^3$
The volume of carbon dioxide produced is $\text{280000 dm}^3 \text{ or } 280 \text{ m}^3$ .

For the volume of carbon dioxide produced from 2100 kg of sodium hydrogen carbonate. The reaction is given as:

Answer

Step-by-step solution

Calculate the molar mass of $\text{NaHCO}_3$ .

Convert the mass of $\text{NaHCO}_3$ from kg to g and then to moles.

Use the stoichiometry of the reaction to find the moles of $\text{CO}_2$ produced.

Calculate the volume of $\text{CO}_2$ at STP.

Convert the volume to $\text{m}^3$ (optional, but often preferred for large volumes).

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More Chemistry Questions

For the volume of carbon dioxide produced from 2100 kg of sodium hydrogen carbonate. The reaction is given as:

Answer

Step-by-step solution

Calculate the molar mass of NaHCO3\text{NaHCO}_3NaHCO3​.

Convert the mass of NaHCO3\text{NaHCO}_3NaHCO3​ from kg to g and then to moles.

Use the stoichiometry of the reaction to find the moles of CO2\text{CO}_2CO2​ produced.

Calculate the volume of CO2\text{CO}_2CO2​ at STP.

Convert the volume to m3\text{m}^3m3 (optional, but often preferred for large volumes).