Question

50g of caco3 is treated with excess of hcl calculate the volume and No. of molecules of co2 liberated​

Answer 1

Hope this helps you...

Answer 2

$caco3 + 2hcl = cacl2 + h2o + co3$

40+12+3×16

$100g \: of \: caco3 + 2 \times 31.5 = 44g \: of \: co2$

50g of caco3 + 7.3 g of HC= ?

73g of CO2 is produced 44g of Ca2

73g of Ca2=?

$x \times 7.3 = 7.3 \times 44$

$x = 44g \: of \: ca2$

Mole:

$n = \binom{w}{g} = \binom{1}{10} = 0.1 \: mole$

No. of particles:

$1 \: mole \: = 6.023 \times {10}^{23}$

0.1 mole =?

$= 6.023 \times {10}^{23} \times 0.1$

$= 6.023 \times {10}^{22} moles$

Volume:

$1 \: mole = 22.4 \: litres$

0.1 mole=?

$= 0.1 \times 22.4$

= 2.24 litres