Question

Amt of 25percent pure caco3 required to obtain 11.2 litr co2 when reacts with H2SO4 ​

Answer 1

answer : option (4) 200g

explanation : when calcium carbonate is heated , gives carbon dioxide and calcium oxide.

i.e., $CaCO_3\rightarrow CaO+CO_2$

here it is clear that one mole of calcium carbonate produced one mole of carbon dioxide.

so, amount of calcium carbonate required to produce 11.2litre CO2 at STP= mole of calcium carbonate in 11.2 litre × molar mass of calcium carbonate

[ you should know, volume of 1 mole of gas at STP/ NTP = 22.4L, molar mass of CaCO3 = 100g/mol ]

= 11.2/22.4 × 100

= 50g

but question said “ find 25% pure CaCO3”

so, 25 % of x = 50g

or, 25 × x/100 = 50g

or, x = 200g

hence, amount of 25% pure CaCO3 is 200g.

Answer 2

Answer: 200 grams

Explanation:

$CaCO_3+H_2SO_4\rightarrow CaSO_4+H_2O+CO_2$

$\text{Number of moles}=\frac{\text{Given volume}}{\text{Molar volume}}$

$\text{Number of moles of carbon dioxide}=\frac{11.2}{22.4}=0.5moles$

According to stoichiometry:

1 mole of $CO_2$ is produced from = 1 mole of $CaCO_3$

0.5 moles of $CO_2$ are  produced from =$\frac{1}{1}\times 0.5=0.5moles$ of $CaCO_3$

Mass of $CaCO_3=[tex]moles\times {\text {Molar mass}}=0.5\times 100=50g$

As  $CaCO_3$ is 25% pure , the amount required will be=$\frac{100}{25}\times 50=200g$

The amount of 25 percent pure  $CaCO_3$ required to obtain 11.2 liter $CO_2$ when reacts with $H_2SO_4$ is 200 g.