Question

Cacium carbonate react with aqueous HCl to give CaCl2 and CO2 according to the reaction given below: CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l). What mass of CaCl2 will be formed when 250ml of 0.76M HCl reacts with 1000g of CaCO3 ? Name the limiting reagent. Calculate the number of moles of CaCl2 formed in the reaction?

Answer 1

it can be solved with two steps :

1) first we will calculate the mass of HCl in 25ml of 0.75M HCl.

2) Now, calculate the mass of CaCO3 by using all information available from balance chemical equation .

step 1 : calculation of mass of HCl in 25 ml of 0.75M HCl .

we know,

Molarity = mass of solute/volume of solution in L

mass of HCl = 0.6844 g

step 2 : calculation of mass of CaCO3 ,

CaCO3 + 2HCl ------> CaCl2 + CO2 + H2O

here we see that,

2 mole of HCl reacts with 1 mole of CaCO3.

so, 2 × 36.5g of HCl reacts with 100g of CaCO3.

so, 73g of HCl reacts with 100g of CaCO3.

so, 1g of HCl reacts with 100/73 g of CaCO3.

so, 0.6844 g of HCl reacts with 100 × 0.6844/73g of CaCO3 = 68.44/73 g = 0.9375g

0.9375g of CaCO3 is required to react completely with 25ml of 0.75M HCl.

Explanation: