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

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Here's the solution to the chemistry problem.

The problem asks for the volume of carbon dioxide produced from 2100 kg of sodium hydrogen carbonate. The reaction is given as: $2NaHCO_3(s) \xrightarrow{heat} Na_2CO_3(s) + CO_2(g) + H_2O(g)$ 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 $\text{dm}^3$.

Step 1: Calculate the molar mass of $\text{NaHCO}_3$. $Molar mass of NaHCO_3 = (23 \times 1) + (1 \times 1) + (12 \times 1) + (16 \times 3)$ $= 23 + 1 + 12 + 48 = 84 g/mol$

Step 2: Convert the mass of $\text{NaHCO}_3$ from kg to g and then to moles. Given mass of $\text{NaHCO}_3 = 2100 kg = 2100 \times 1000 g = 2.1 \times 10^6 \text{ g}$ $Moles of NaHCO_3 = \frac{Mass}{Molar mass} = \frac{2.1 \times 10^6 g}{84 g/mol}$ $= 25000 mol$

Step 3: 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 $\text{CO}_2$. $Moles of CO_2 = Moles of NaHCO_3 \times \frac{1 mol CO_2}{2 mol NaHCO_3}$ $= 25000 mol \times \frac{1}{2} = 12500 mol$

Step 4: Calculate the volume of $\text{CO}_2$ at STP. At STP, 1 mole of gas occupies 22.4 $\text{dm}^3$. $Volume of CO_2 = Moles of CO_2 \times 22.4 dm^3/mol$ $= 12500 mol \times 22.4 dm^3/mol$ $= 280000 dm^3$

Step 5: Convert the volume to $\text{m}^3$ (optional, but often preferred for large volumes). Since $1 m^3 = 1000 dm^3$: $Volume of CO_2 = \frac{280000 dm^3}{1000 dm^3/m^3} = 280 m^3$

The volume of carbon dioxide produced is $\boxed{280000 dm^3 or 280 m^3}$.

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