Volume of (CO\_2) obtained by the complete decomposition of... | Chemistry | SolveeIt

Volume of $CO_2$ obtained by the complete decomposition of $9.85\ gm \ BaCO_3$ is:

$2.24 \ lit.$

$1.12\ lit.$

$0.84\ lit.$

$0.56\ lit.$

Answer

$1.12\ lit.$

Explanation

Molar mass of Ba = 137.33 g/mol

Molar mass of C = 12.01 g/mol

Molar mass of O = 16.00 g/mol

Molar mass of BaCO3 = 137.33 + 12.01 + 3 x 16.00 = 197.34 g/mol

Moles of BaCO3 = $\frac {Mass }{Molar \ mass }$

Moles of BaCO3 = $\frac {9.85\ g }{ 197.34\ g/mol}$ ≈ 0.05 moles

Balanced chemical equation for the decomposition of BaCO3

$BaCO_3 → BaO + CO_2$

The stoichiometry tells us that one mole of BaCO3 produces one mole of CO2.

Moles of CO2 = Moles of BaCO3 ≈ 0.05 moles

Now, use the ideal gas law to find the volume of CO2 at standard temperature and pressure (STP):

The molar volume of an ideal gas at STP is 22.4 liters/mol.

Volume of CO2 = Moles of CO2 x Molar volume at STP

Volume of CO2 = 0.05 moles x 22.4 liters/mole = 1.12 liters

So, the correct option is (B): $1.12\ lit.$

Chemistry Question on Some basic concepts of chemistry