Caco3 + 2HCl = CaCl2 + Co2 + H2O What is the amount of CaCO3 to process thoroughly with 25 ml 0.75 M HCl?
Answers
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.