after heating x gram of sodium bicarbonate 5.6 litre of co2 is liberated at STP then the value of x is
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Answer:
The reaction is CaCO
3
→CaO+CO
2
.
One mole of calcium carbonate forms one mole of carbon dioxide.
At STP, 5.6 L of carbon dioxide corresponds to 0.25 moles.
They will be obtained from 0.25 moles of calcium carbonate.
Molar mass of CaCO
3
=100g
0.25 moles of calcium carbonate (molecular weight 100 g/mole) corresponds to 25 g.
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