Respuesta :
Answer: The mass of [tex]CO_2[/tex] produced is 0.44 g
Explanation:
- For calcium carbonate:
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
Given mass of calcium carbonate = 10 g
Molar mass of calcium carbonate = 100 g/mol
Plugging values in equation 1:
[tex]\text{Moles of calcium carbonate}=\frac{10g}{100g/mol}=0.1 mol[/tex]
- For HCl:
The formula used to calculate molarity:
[tex]\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}}[/tex] .....(2)
Molarity of HCl = [tex]0.2mol/dm^3=0.2mol/L[/tex] (Conversion factor: [tex]1L=1dm^3[/tex]
Volume of solution = [tex]100cm^3=100mL[/tex] (Conversion factor: [tex]1mL=1cm^3[/tex]
Putting values in equation 2, we get:
[tex]0.2=\frac{\text{Moles of HCl}\times 1000}{100}\\\\\text{Moles of HCl}=\frac{0.2\times 100}{1000}=0.02mol[/tex]
For the given chemical reaction:
[tex]CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2O(l)+CO_2(g)[/tex]
By stoichiometry of the reaction:
If 2 moles of HCl reacts with 1 mole of calcium carbonate
So, 0.02 moles of HCl will react with = [tex]\frac{1}{2}\times 0.02=0.01mol[/tex] of calcium carbonate
As the given amount of calcium carbonate is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.
Thus, HCl is considered a limiting reagent because it limits the formation of the product.
By the stoichiometry of the reaction:
If 2 moles of [tex]HCl[/tex] produces 1 mole of [tex]CO_2[/tex]
So, 0.02 moles of [tex]HCl[/tex] will produce = [tex]\frac{1}{2}\times 0.02=0.01mol[/tex] of [tex]CO_2[/tex]
We know, molar mass of [tex]CO_2[/tex] = 44 g/mol
Putting values in equation 1, we get:
[tex]\text{Mass of }CO_2=(0.01mol\times 44g/mol)=0.44g[/tex]
Hence, the mass of [tex]CO_2[/tex] produced is 0.44 g
