How many moles of sulfuric acid is required to dissolve 3 gram of $M g C O_{3}$ ?

9 m o l

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0.036 m o l

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\frac{4}{5} m o l

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\frac{11}{2} m o l

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Solution

The correct option is B $0.036 m o l$
We have

$H_{2} S O_{4} + M g C O_{3} \to H_{2} C O_{3} + M g S O_{4}$

so, for 1 mole of $M g S O_{4}$ we require 1 mole of $H_{2} S O_{4}$ .

moles of $M g C O_{3}$ = $\frac{3}{84} = 0.0357 m o l e s$

hence, moles of $H_{2} S O_{4}$ require = $0.036 m o l e s$ .

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How many moles of sulfuric acid is required to dissolve 3 gram of MgCO3?

The correct option is B 0.036molWe haveH2SO4+MgCO3→H2CO3+MgSO4so, for 1 mole of MgSO4 we require 1 mole of H2SO4.moles of MgCO3 = 384=0.0357moleshence, moles of H2SO4 require = 0.036moles.

How many moles of sulfuric acid is required to dissolve 3 gram of $M g C O_{3}$ ?