Question

how many molecules of CO2 are formed when one milligram of 100% pure CaCO3 is treated with excess hydrochloric acid?

please please don't spam it's urgent​

Answer 1

Explanation:

ANSWER

100 g

CaCO

3

+2HCl→CaCl

2

+H

2

O+

1 mol

CO

2

∵100 g CaCO

3

gives, molecules of CO

2

=6.022×10

23

∴1×10

−3

g CaCO

3

gives molecules of CO

2

=

100

6.022×10

23

×1×10

−3

=6.022×10

18

.

Answer 2

Answer:

Caco3 + HCl _ Cacl2 + H2O + Co2

100g 1mol

100g of Caco3 gives 1mol of Co2 = 6.022 × 10^23

= 6.022 × 10^23 × 1 × 10^-3 /100

Ans = 6.022× 10^18