Question

Example 1.6 How many grams of CO2 can be produced by thermally decomposing 8.8 moles of ZnCO3(s)?​

Answer 1

Answer:

387.29 g of $CO_{2}$ can be produced by thermally decomposing 8.8 moles of $ZnCO_{3}$.

Explanation:

Zinc carbonate decomposes to give Zinc oxide and carbon dioxide.

The balanced chemical equation for the reaction can be written as:

$ZnCO_{3}$ → $ZnO + CO_{2}$

From the balanced equation:

1 mole $ZnCO_{3}$ → 1 mole $CO_{2}$

8.8 mole $ZnCO_{3}$ → 8.8 mole $CO_{2}$

Number of moles = given mass ÷ molar mass

We know:

Number of moles = 8.8 moles

Molar mass of $CO_{2}$ = 44.01 g/mol

Substituting in the above equation:

Given mass = 8.8 × 44.01

                    = 387.29 g

Answer 2

Answer:

387.2g grams of CO2 can be produced by thermally decomposing 8.8 moles of ZnCO₃.

Explanation:

the chemical equation of decomposition of ZnCO₃ is given by,

           ZnCO₃  →  ZnO + CO₂

So, one mole of ZnCO₃ gives one mole of CO₂.

⇒   8.8 moles of ZnCO₃ gives 8.8 moles of CO₂.

mass of a molecule is given by,

      mass = number of moles × molar mass

molar mass of CO₂ = 12 + 2×16 = 44g/mol.

         mass of CO₂ = 8.8 × 44 = 387.2g

Hence, 387.2g of CO₂ is produced.