Respuesta :
Answer: The mass of [tex]CaCl_2[/tex] produced is, 15.2 grams.
Explanation : Given,
Mass of [tex]CaCO_3[/tex] = 30.0 g
Mass of [tex]HCl[/tex] = 10.0 g
Molar mass of [tex]CaCO_3[/tex] = 100 g/mol
Molar mass of [tex]HCl[/tex] = 36.5 g/mol
First we have to calculate the moles of [tex]CaCO_3[/tex] and [tex]HCl[/tex].
[tex]\text{Moles of }CaCO_3=\frac{\text{Given mass }CaCO_3}{\text{Molar mass }CaCO_3}[/tex]
[tex]\text{Moles of }CaCO_3=\frac{\text{Given mass }CaCO_3}{\text{Molar mass }CaCO_3}=\frac{30.0g}{100g/mol}=0.300mol[/tex]
and,
[tex]\text{Moles of }HCl=\frac{\text{Given mass }HCl}{\text{Molar mass }HCl}[/tex]
[tex]\text{Moles of }HCl=\frac{\text{Given mass }HCl}{\text{Molar mass }HCl}=\frac{10.0g}{36.5g/mol}=0.274mol[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
[tex]CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2O(l)+CO_2(g)[/tex]
From the balanced reaction we conclude that
As, 2 mole of [tex]HCl[/tex] react with 1 mole of [tex]CaCO_3[/tex]
So, 0.274 moles of [tex]HCl[/tex] react with [tex]\frac{0.274}{2}=0.137[/tex] moles of [tex]CaCO_3[/tex]
From this we conclude that, [tex]CaCO_3[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]HCl[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]CaCl_2[/tex]
From the reaction, we conclude that
As, 2 mole of [tex]HCl[/tex] react to give 1 mole of [tex]CaCl_2[/tex]
So, 0.274 mole of [tex]HCl[/tex] react to give [tex]\frac{0.274}{2}=0.137[/tex] mole of [tex]CaCl_2[/tex]
Now we have to calculate the mass of [tex]CaCl_2[/tex]
[tex]\text{ Mass of }CaCl_2=\text{ Moles of }CaCl_2\times \text{ Molar mass of }CaCl_2[/tex]
Molar mass of [tex]CaCl_2[/tex] = 110.98 g/mole
[tex]\text{ Mass of }CaCl_2=(0.137moles)\times (110.98g/mole)=15.2g[/tex]
Therefore, the mass of [tex]CaCl_2[/tex] produced is, 15.2 grams.
