Calculate the volume of CO2 produced at STP on the decomposition of 20g MgCO3 which is 42% pure. please it's urgent
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Answer:
44gm=22.4l
8.8gm=4.48l
Explanation:
Decomposition of CaCO3
⇒CaCO3=CaO+CO2
⇒1mole=1mole+1mole
⇒100gm=56gm+44gm
∴For 20gm of CaCO3
⇒20gm=11.5gm+8.8gm
but qn is to find the volume of CO2
1mole=22.4l
44gm=22.4l
8.8gm=4.48l
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