Question

How much of 80% pure CaCO3 will be required to produce 44.8 L of CO2 at STP?

Answer 1

Hey dear,

● Answer- 250 g

● Solution-
Reaction is
CaCO3 ---> CaO + CO2

According to ideal gas eqn-
n = V / V0 where
V0 = volume at STP
n = 44.8 / 22.4
n = 2

No of moles of CO2 = No of moles of CaCO3

Weight of 80% pure CaCO3 = M × n / purity
Weight of 80% pure CaCO3 =
100/80 × 100 × 2
Weight of 80% pure CaCO3 =
250 g

Hope that was useful...

Answer 2

Answer:- 250 g

Solution:- The balanced equation for the decomposition of calcium carbonate to give carbon dioxide is:

$CaCO_3\rightarrow CaO+CO_2$

From above equation, there is 1:1 mol ratio between calcium carbonate and carbon dioxide, so the moles of calcium carbonate required will exactly be equal to the moles of carbon dioxide formed.

moles of carbon dioxide are calculated from its given STP volume. For STP, we know that the volume of 1 mol of the gas is 22.4 L. So:

$44.8LCO_2(\frac{1mol}{22.4L)})$

= $2molCO_2$

Now, the moles of calcium carbonate are calculated as:

$2molCO_2(\frac{1molCaCO_3}{1molCO_2})$

= $2molCaCO_3$

Molar mass of calcium carbonate is 100 gram per mol. Multiply the moles by molar mass to get the grams.

$2molCaCO_3(\frac{100g}{1mol})$

= 200g

The sample is 80% pure. It means 100 g sample has only 80 g calcium carbonate. How many grams of the sample would have 200 g of calcium carbonate. It is calculated as:

$200gCaCO_3(\frac{100gsample}{80gCaCO_3})$

= 250g sample

Hence, 250 g of 80% pure calcium carbonate will be required.