Question

What volume of CO2 gas at NTP can be obtained from 2 kg of Caco3​

Answer 1

448 L

Explanation:

The balanced reaction is

CaCO₃ → CaO + CO₂

1 mole of CaCO₃ gives 1 mole of  CO₂

100 g of CaCO₃  gives 22.4 L of  CO₂

At NTP V = 22.4 L

2000 g of CaCO₃ will give x L of CO₂

x = 22.4 × 2000/100  

  = 20 × 22.4

  = 448 L

Answer 2

Answer:

448 L of $CO_{2}$ gas at NTP can be obtained from 2 kg of $CaCO_{3}$.

Explanation:

The balanced chemical equation for the decomposition of $CaCO_{3}$ is:

$CaCO_{3}$ → $CaO + CO_{2}$

Given that mass of $CaCO_{3}$ = 2 kg = 2000 g

Molar mass of $CaCO_{3}$ = (1×40) + (1×12) + (3×16)

                                    = 100 g/mol

Number of moles of $CaCO_{3}$ = given mass ÷ molar mass

                             = $\frac{2000}{100}$

                             = 20 moles

∴Number of moles of $CO_{2}$ = 20 moles

1  mole of $CO_{2}$ occupies 22.4 L at NTP

∴ 20 moles of $CO_{2}$ occupy 22.4 × 20

                                              = 448 L at NTP