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Chemistry

4.2 grams of magnesium carbonate is decomposed by heating according to the question.

MgCO3 ⟶ MgO + CO2

Calculate the following

(a) Volume of carbon dioxide obtained at STP.

(b) Mass of MgO formed

[Atomic weight : Mg = 24, C = 12, O = 16]

Stoichiometry

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Answer

MgCO3⟶MgO+CO224+12+3(16) g=24+16[12+2(16)]=84=4012+3244 g\begin{matrix} \text{MgCO}3 & \longrightarrow & \text{MgO} & + & \text{CO}2 \ 24 + 12 + 3(16) \text{ g} & & = 24 + 16 & & [ 12 + 2(16)] \ = 84 & & = 40 & & 12 + 32 \ & & & & 44 \text{ g} \ \end{matrix}

(a) Number of moles of MgCO3 = Given weightmolecular weight\dfrac{\text{Given weight}}{\text{molecular weight}} = 4.284\dfrac{4.2}{84} = 0.05 moles

1 mole of MgCO3 gives 1 mole of CO2

Hence, 0.005 mole of MgCO3 gives 0.05 mole of CO2

(a) Volume of carbon dioxide at STP = no. of moles of carbon dioxide x volume in litres
= 0.05 x 22.4 = 1.12 L

Hence, volume of carbon dioxide obtained at STP = 1.12 L

(b) 84 grams of MgCO3 produce 40 grams of MgO

∴ 4.2 g will produce = 4084\dfrac{40}{84} x 4.2 = 2 g

Hence, mass of MgO formed = 2g

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