Question

Calcium carbonate reacts with HCl to give CaCl2
and Co, according to the reaction:
CaCO3(s) + 2HCl(aq) → CaCl (aq) + CO2(g) +
H2O(1) What mass of 20% impure CaCO3 is
required to react completely with 250 mL of 0.50
M HCI?
(1) 6.25 g
(2) 5 g
(3) 4.75 g
(4) 7.8 g​

Answer 1

Explanation:

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

0.75=\frac{mass\:of\:HCl}{25ml}\times\:10000.75=

25ml

massofHCl

×1000

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 .

If u like my answer than plzz follow me and mark me as BRAINLIST