Test Index

JAMB Chemistry 2020 Paper

© africaexams.com

Question : 17

Total: 40

The number of molecules of Carbon(iv)Oxide produced when 10.0g of CaCO

_3

is treated with 0.2dm

^3

of 1 Mole of HCL in the equation

1.00 X 10 $^{23}$
6.02 X 10 $^{23}$
6.02 X 10 $^{22}$
6.02 X 10 $^{24}$

Solution:

1 mole of CaCO $_3$ = 100g or 6.02 X 10 $^{23}$

liberated 1 mole of CO $_2$ or 44g or 6.02 X 10 $^{23}$

:100g = 6.02 X 10 $^{23}$

10g of CaCO $_3$ = x

cross multiply

[10g X 6.02 X 10 $^{23}$ ] / 100g = x

⇒ x = 6.02 X 10 $^{22}$

Since

6.02 X 10 $^{23}$ of CaCO $_3$ liberated 6.02 X 10 $^{23}$ of CO $_2$

⇒ 6.02 X 10 $^{22}$ of CaCO $_3$ will liberate 6.02 X 10 $^{22}$ of CO $_2$

© africaexams.com

Go to Question:

Prev Question

Next Question