Question

Volume of Co2 obtained at STP by the complete decomposition of 9.85gm of BaCO3 is ______

Answer 1

Carbonates are decomposed like:

Metal Carbonate ⇒ Metal oxide + Carbon Dioxide

So :

BaCO3 ⇒ BaO + CO2

Let's find moles of BaCO3

moles = mass/ Mr

= 9.85 / 197

= 0.05 mol

Since the 1 mole of each is present in the equation, CO2 also has 0.05 mol

Volume = mole × Mr volume(24)

= 0.05 × 24

= 1.2 $dm^{3}$

Therefore the volume of CO2 is 1.2 dm^{3}

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