
The mass of sulphuric acid needed for dissolving 3g magnesium carbonate is:
(a) 3.5 g
(b) 7 g
(c) 1.7 g
(d) 17.0 g
Answer
595.8k+ views
Hint: Magnesium carbonate is also called magnesite. When an insoluble metal oxide, metal hydroxides and metal carbonates react with a dilute acid, it forms a soluble salt, water and carbon dioxide gas is evolved.
Complete step-by-step answer:
Magnesium carbonate does not dissolve in sulphuric acid, instead it reacts with sulphuric acid to form a soluble salt of magnesium sulphate, followed by evolution of carbon dioxide gas and formation of water molecules. The chemical reaction can be given as:
\[{{H}_{2}}S{{O}_{4}}+MgC{{O}_{3}}\to MgS{{O}_{4}}+{{H}_{2}}O+C{{O}_{2}}\]
Now, let us calculate the molar mass of \[MgC{{O}_{3}}\],
We know the Atomic masses of:
Mg = 24.3050 g/mol
C = 12.0107 g/mol
O = 15.9994 g/mol
Molar mass of \[MgC{{O}_{3}}\]= 24.305 + 12.0107 + (15.9994 × 3)
= 84 g/mol
Now, let us calculate the molar mass of \[{{H}_{2}}S{{O}_{4}}\]
We know the Atomic masses of:
H = 1 g/mol
S = 32.066 g/mol
O = 15.9994 g/mol
Molar mass of \[{{H}_{2}}S{{O}_{4}}\]= (1.00794 × 2) + 32.065 + (15.9994 × 4)
= 98 g/mol
If we write in terms of mass then, 84g of \[MgC{{O}_{3}}\]reacts with 98g of \[{{H}_{2}}S{{O}_{4}}\].
Therefore, using unitary method we can calculate that,
1g of \[MgC{{O}_{3}}\]is equals to \[\dfrac{1}{84}mol\]
Then, 3g of \[MgC{{O}_{3}}\] will be equal to \[3\times \dfrac{1}{84}mol=0.036\,mol\]
We can see in the reaction that 1 mole of \[MgC{{O}_{3}}\] reacts with 1 mole of \[{{H}_{2}}S{{O}_{4}}\].
Similarly, 0.036 moles of \[MgC{{O}_{3}}\] will react with 0.036 moles of \[{{H}_{2}}S{{O}_{4}}\].
Finally, \[0.036\,\times 98.1=3.5g\]of sulphuric acid.
So, option (a) is correct.
Additional Information:
Magnesite is the primary source for production of magnesium and its compound.
It is also used as raw material in iron-steel, cement, glass, sugar, lime and paper industries as well as paint and ink industry, pharmaceutical industry as an antacid.
Note: Leaching of magnesite with sulphuric acid is of great importance from industrial point of view, as it is used for obtaining pure magnesium sulphate. The above reaction is used to obtain pure heptahydrate colorless crystals of magnesium sulphate.
Complete step-by-step answer:
Magnesium carbonate does not dissolve in sulphuric acid, instead it reacts with sulphuric acid to form a soluble salt of magnesium sulphate, followed by evolution of carbon dioxide gas and formation of water molecules. The chemical reaction can be given as:
\[{{H}_{2}}S{{O}_{4}}+MgC{{O}_{3}}\to MgS{{O}_{4}}+{{H}_{2}}O+C{{O}_{2}}\]
Now, let us calculate the molar mass of \[MgC{{O}_{3}}\],
We know the Atomic masses of:
Mg = 24.3050 g/mol
C = 12.0107 g/mol
O = 15.9994 g/mol
Molar mass of \[MgC{{O}_{3}}\]= 24.305 + 12.0107 + (15.9994 × 3)
= 84 g/mol
Now, let us calculate the molar mass of \[{{H}_{2}}S{{O}_{4}}\]
We know the Atomic masses of:
H = 1 g/mol
S = 32.066 g/mol
O = 15.9994 g/mol
Molar mass of \[{{H}_{2}}S{{O}_{4}}\]= (1.00794 × 2) + 32.065 + (15.9994 × 4)
= 98 g/mol
If we write in terms of mass then, 84g of \[MgC{{O}_{3}}\]reacts with 98g of \[{{H}_{2}}S{{O}_{4}}\].
Therefore, using unitary method we can calculate that,
1g of \[MgC{{O}_{3}}\]is equals to \[\dfrac{1}{84}mol\]
Then, 3g of \[MgC{{O}_{3}}\] will be equal to \[3\times \dfrac{1}{84}mol=0.036\,mol\]
We can see in the reaction that 1 mole of \[MgC{{O}_{3}}\] reacts with 1 mole of \[{{H}_{2}}S{{O}_{4}}\].
Similarly, 0.036 moles of \[MgC{{O}_{3}}\] will react with 0.036 moles of \[{{H}_{2}}S{{O}_{4}}\].
Finally, \[0.036\,\times 98.1=3.5g\]of sulphuric acid.
So, option (a) is correct.
Additional Information:
Magnesite is the primary source for production of magnesium and its compound.
It is also used as raw material in iron-steel, cement, glass, sugar, lime and paper industries as well as paint and ink industry, pharmaceutical industry as an antacid.
Note: Leaching of magnesite with sulphuric acid is of great importance from industrial point of view, as it is used for obtaining pure magnesium sulphate. The above reaction is used to obtain pure heptahydrate colorless crystals of magnesium sulphate.
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