Question

Mass of 50% pure CaCO3 required to produce 5.6 L of
CO2(g) at S.T.P. is​

Answer 1

Answer:

The STP is 400g

Explanation:

• Calcium Mass = 40.
• Carbon mass = 12
• Oxygen Mass = 16.
• Now we have 1 mole of Calcium and Carbon, But Three moles of Oxygen. So we have to multiply the Mass of Oxygen with 3.
• I.e. 16 × 3 = 48.
• Now by adding all these, we can get the Formula mass of CaCO3 ( Calcium Carbonate ).
• Formula Mass = 40 + 12 + 48 = 100 Grams.
• Therefore, The mass of Calcium Carbonate is 100 grams .
• The reaction is CaCO3 →CaO+CO2 .
• According to ideal gas eqn-
• n = V / V0 where
• V0 = volume at STP
• n = 5.6 / 2.8
• n = 2
• No of moles of CO2 = No of moles of CaCO3
• Weight of 50% pure CaCO3 = M × n / purity
• Weight of 50% pure CaCO3 =
• 100/50 × 100 × 2
• Weight of 50% pure CaCO3 =
• 400 g
• One mole of calcium carbonate forms one mole of carbon dioxide.
• At STP, 5.6 L of carbon dioxide corresponds to 0.25 moles.
• They will be obtained from 0.25 moles of calcium carbonate.
• Molar mass of CaCO3​ =100g
• 0.25 moles of calcium carbonate (molecular weight 100 g/mole) corresponds to 25 g.

Reference Link

• https://brainly.in/question/7322783
• https://brainly.in/question/9861452

Answer 2

The answer is 49.94 grams

GIVEN

Amount of CO₂ formed = 5.6l

Purity of CaCO₃ = 50%

TO FIND

Mass of CaCO₃

SOLUTION

We can simply solve the above problem as follows;

CaCO₃ on decomposition undergoes following reaction;

CaCO₃ ----------> CaO + CO₂

We can observe that 1 mole of CaCO₃ produces 22.4litres of CO₂ at STP

1 Mole of CaCO₃ = Molar mass of CaCO₃ = 100 grams

Mass of CaCO₃ producing 1 litre of CO₂ = 100/22.4 grams

Mass of CaCO₃ producing 5.6 litres of CO₂ =

$\frac{100}{22.4} \times 5.6 = 24.97grams$

It is given that sample is only 50% pure.

So The weight of the sample should be double the calculated mass of CaCO₃

Therefore,

Mass of the sample = 24.97 × 2 = 49.94 grams

Hence, The answer is 49.94 grams

#Spj2