4 moles of MgCO3 is reacted with 6 moles of HCl solution. Find the volume of CO2 gas produced at STP. MgCO3+2HCl→MgCl2+CO2+H2O
A
67.2 lit
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B
89.6 lit
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C
76.3 lit
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D
22.4 lit
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Solution
The correct option is A 67.2 lit Considering the equation which is already balanced: MgCO3+2HCl→MgCl2+CO2+H2O , Limiting reagent here is HCl Therefore, 6 moles of HCl will produce 3 moles of CO2 Volume occupied by 3 moles =3×22.4=67.2L