What is the weight of H_2SO_4 (in g) required | vedclass

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(C) The chemical equation for the reaction is:$MgCO_3 + H_2SO_4 ightarrow MgSO_4 + CO_2 + H_2O$Calculate the molar masses:$MgCO_3 = 24 + 12 + (3 	i

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$3.50$

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(C) The chemical equation for the reaction is:$MgCO_3 + H_2SO_4 ightarrow MgSO_4 + CO_2 + H_2O$Calculate the molar masses:$MgCO_3 = 24 + 12 + (3 	imes 16) = 84 \ g/mol$$H_2SO_4 = (2 	imes 1) + 32 + (4 	imes 16) = 98 \ g/mol$From the stoichiometry,$84 \ g$ of $MgCO_3$ reacts with $98 \ g$ of $H_2SO_4$.Therefore,for $3 \ g$ of $MgCO_3$,the mass of $H_2SO_4$ required is:$\frac{98 \ g}{84 \ g} 	imes 3 \ g = 3.5 \ g$.

What is the weight of $H_2SO_4$ (in $g$) required to completely dissolve $3 \ g$ of magnesium carbonate?

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