Question

Calculate the mass, volume and no. of molecules of carbon dioxide liberated when 1272 g of

sodium carbonate reacts with 1460 g dilute hydrochloric acid at STP.
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Answer 1

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Answer 2

Given:

Mass of Sodium carbonate = 1272 gm

Mass of Hydrochloric Acid = 1460 gm

To Find:

The mass, volume and no. of molecules of carbon dioxide liberated on this reaction.

Calculation:

- The chemical reaction is:

Na2CO3 + 2 HCl → 2 NaCl + CO2 + H2O

- 1 mole of Na2CO3 reacts with 2 moles of HCl to produce 1 mole of CO2.

⇒ 106 gm of Na2CO3 reacts with 73 gm of HCl to produce 44 gm CO2.

⇒ 1272 gm of Na2CO3 reacts with 876 gm of HCl to produce 528 gm CO2.

- No of moles of CO2 = 528/44 = 12

- The volume of 1 mole of CO2 = 22.4 L

⇒ The volume of 12 moles of CO2 = 22.4 × 12

⇒ V = 268.8 L

- No of molecules in 1 mole CO2= 6.022 × 10²³

⇒ No of molecules in 12 moles of CO2 = 12 × 6.022 × 10²³

⇒ N = 72.264 × 10²³

- So, the mass of CO2 is 528 gm, the volume is 268.8 L and the no of molecules is 72.264 × 10²³.