Respuesta :
Answer: a) 18.3 grams
b) [tex]CaCO_3[/tex] is the excess reagent and 16.5g of [tex]CaCO_3[/tex] will remain after the reaction is complete.
Explanation:
[tex]CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2[/tex]
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
[tex]\text{Number of moles of calcium carbonate}=\frac{26g}{100g/mol}=0.26moles[/tex]
[tex]\text{Number of moles of hydrochloric acid}=\frac{12g}{36.5g/mol}=0.33moles[/tex]
According to stoichiometry:
2 mole of [tex]HCl[/tex] react with 1 mole of [tex]CaCO_3[/tex]
0.33 moles of [tex]HCl[/tex] will react with=[tex]\frac{1}{2}\times 0.33=0.165moles[/tex] of [tex]CaCO_3[/tex]
Thus [tex]HCl[/tex] is the limiting reagent as it limits the formation of product and [tex]CaCO_3[/tex] is the excess reagent.
2 moles of [tex]HCl[/tex] produce = 1 mole of [tex]CaCl_2[/tex]
0.33 moles of [tex]HCl[/tex] produce=[tex]\frac{1}{2}\times 0.33=0.165moles[/tex] of [tex]CaCl_2[/tex]
Mass of [tex]CaCl_2=moles\times {\text{Molar Mass}}=0.165\times 111=18.3g[/tex]
As 0.165 moles of [tex]CaCO_3[/tex] are used and (0.33-0.165)=0.165 moles of [tex]CaCO_3[/tex] are left unused.
Mass of [tex]CaCO_3[/tex] left unreacted =[tex]moles\times {\text {Molar mass}}=0.165\times 100=16.5g[/tex]
Thus 18.3 g of [tex]CaCl_2[/tex] are produced. [tex]CaCO_3[/tex] is the excess reagent and 16.5g of [tex]CaCO_3[/tex] will remain after the reaction is complete.
