How many grams of sulphuric acid is required for 1.3 grams of magnesium carbonate ?
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let’s see the equation.
H2SO4+MgCO3 ------> MgSO4+H2CO3
It is evident that 1 mol of sulphuric acid is required to dissolve 1 mol MgCO3
Molar mass of Magnesium Carbonate = 84.3g
Therefore,
3g of MgCO3 = 3/ 84.3 mol = 0.036 mol
Therefore, 0.036 mol of Sulphuric acid is required to dissolve 0.036 mol of MgCO3
Or, 0.036 x 98.1=3.5g of H2SO4 are required.
H2SO4+MgCO3 ------> MgSO4+H2CO3
It is evident that 1 mol of sulphuric acid is required to dissolve 1 mol MgCO3
Molar mass of Magnesium Carbonate = 84.3g
Therefore,
3g of MgCO3 = 3/ 84.3 mol = 0.036 mol
Therefore, 0.036 mol of Sulphuric acid is required to dissolve 0.036 mol of MgCO3
Or, 0.036 x 98.1=3.5g of H2SO4 are required.
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