Q.

The weight of MgCO3 required for the preparation of 12g of MgSO4 using H2SO4 is

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a

8.4g

b

4.2g

c

16.8g

d

12.6g

answer is A.

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Detailed Solution

MgCO3 +H2SO4→MgSO4+H2O+CO2 1 mole of MgCO3 -3 moles of MgSO4 84 gms of MgCO3-120 gms of MgSO4 X gms of MgCO3 -12 gms of MgSO4 X=12120×84 =8.4 gmsX = 8.4g
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