Some Basic Concepts Of Chemistry

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Q1.

When 7.3 g of magnesium bicarbonate (Mg(HCO₃)₂) is decomposed by heating, how much carbon dioxide gas (in mL) is released at STP?

Explanation:

Hint: The reaction is $Mg {({HCO}_{3})}_{2} \to {MgCO}_{3} + H_{2} O + {CO}_{2}$

Step 1:

The reaction is as follows:

Mg(HCO₃)₂ $\to$ MgCO₃ + H₂O +CO₂

No. of moles of Mg(HCO₃)₂ in 7.3 g

= \frac{7.3}{moles of 1 Mg {({HCO}_{3})}_{2}} = \frac{7.3}{146} = 0.05 moles

0.05 moles of Mg(HCO₃)₂ will generate 0.05 moles of CO₂

Step 2:

At STP, 1 mole of gas is 22.4 L or 22400 mL

Volume of

{CO}_{2} = \frac{5}{100} \times 22400 = 1120 mL

The decomposition temperature for MgCO₃ is 350 degree, and Here STP is written.