50 grams of 80%caco3 is heated to calculate the volume of co2 produced at Ntp
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50g of 80% CaCO3 implies that pure calcium carbonate is just
80/100*50g=40g
40g calcium carbonate implies 0.4mol
According to reaction
No. Of moles of carbon dioxide = 0.4mol
Volume at NTP
=> 0.4*22.4L
Which is equal to 8.96L
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