The weight of sulphuric acid needed for dissolving $3 g m$ magnesium carbonate is:

3.5 g m

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7.0 g m

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1.7 g m

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17.0 g m

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Solution

The correct option is A $3.5 g m$
The balanced equation is:
$H_{2} S O_{4} + M g C O_{3} \to M g S O_{4} + H_{2} C O_{3}$
$98 g$ $84 g$
Since, $84 g M g C O_{3} r e q u i r e s = 98 g H_{2} S O_{4}$

Therefore, $3 g$ of $M g C O_{3}$ = $\frac{98 \times 3}{84} = 3.5 g$

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Similar questions

Q. The minimum mass of sulphuric acid needed to dissolve 3 gm of magnesium carbonate is:

Q. The mass of sulphuric acid needed for dissolving

3 g

magnesium carbonate is:

What weight of sulphuric acid will be required to dissolve 3 grams of magnesium carbonate?
[Mg = 24, C = 12, O = 16]
MgCO₃ + H₂SO₄ $\overset{}{\to}$ MgSO₄ + H₂O + CO₂

The burning of magnesium strip resutls in the formation of magnesium oxide. What wil happen if we dissolve this magnesium oxide in sulphuric acid?

Write the balanced equations for the following reactions, and add the state symbols :

(a) Magnesium carbonate reacts with hydrochloric acid to produce magnesium chloride, carbon dioxide and water.

(b) Sodium hydroxide reacts with sulphuric acid to produce sulphate and water

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The weight of sulphuric acid needed for dissolving 3gm magnesium carbonate is:

The correct option is A 3.5gmThe balanced equation is:H2SO4+MgCO3→MgSO4+H2CO3 98 g 84 g Since, 84 g MgCO3 requires=98g H2SO4Therefore, 3 g of MgCO3 = 98×384=3.5 g

The weight of sulphuric acid needed for dissolving $3 g m$ magnesium carbonate is:

The correct option is A $3.5 g m$
The balanced equation is:
$H_{2} S O_{4} + M g C O_{3} \to M g S O_{4} + H_{2} C O_{3}$
$98 g$ $84 g$
Since, $84 g M g C O_{3} r e q u i r e s = 98 g H_{2} S O_{4}$

Therefore, $3 g$ of $M g C O_{3}$ = $\frac{98 \times 3}{84} = 3.5 g$