Respuesta :
This is an incomplete question, here is a complete question.
Identify the limiting reactant when 65.14 g of CaCl₂, reacts with 74.68 g of Na₂CO₃ to produce CaCO₃, and NaCl.
Answer: The limiting reactant is, [tex]CaCl_2[/tex]
Explanation : Given,
Mass of [tex]CaCl_2[/tex] = 65.14 g
Mass of [tex]Na_2CO_3[/tex] = 74.68 g
Molar mass of [tex]CaCl_2[/tex] = 111 g/mol
Molar mass of [tex]Na_2CO_3[/tex] = 106 g/mol
First we have to calculate the moles of [tex]CaCl_2[/tex] and [tex]Na_2CO_3[/tex].
[tex]\text{Moles of }CaCl_2=\frac{\text{Given mass }CaCl_2}{\text{Molar mass }CaCl_2}[/tex]
[tex]\text{Moles of }CaCl_2=\frac{65.14g}{111g/mol}=0.587mol[/tex]
and,
[tex]\text{Moles of }Na_2CO_3=\frac{\text{Given mass }Na_2CO_3}{\text{Molar mass }Na_2CO_3}[/tex]
[tex]\text{Moles of }Na_2CO_3=\frac{74.68g}{106g/mol}=0.704mol[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
[tex]CaCl_2+Na_2CO_3\rightarrow CaCO_3+2NaCl[/tex]
From the balanced reaction we conclude that
As, 1 mole of [tex]CaCl_2[/tex] react with 1 mole of [tex]Na_2CO_3[/tex]
So, 0.587 mole of [tex]CaCl_2[/tex] react with 0.587 mole of [tex]Na_2CO_3[/tex]
From this we conclude that, [tex]Na_2CO_3[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]CaCl_2[/tex] is a limiting reagent and it limits the formation of product.
Hence, the limiting reactant is, [tex]CaCl_2[/tex]
