Question

8.4 g mgco3 on heating leaves behind a residue weighing 4.0 g carbon dioxide released into the atmosphere at s.T.P. Will be -

Answer 1

Answer:

Stoichiometric Calculation:

First of all let's find the number of moles of MgCO₃:

$\footnotesize\implies n_{(MgCO_3)} = \dfrac{Mass}{Molar \: mass}$

$\footnotesize\implies n_{(MgCO_3)} = \dfrac{8.4}{24 + 12 + (16 \times 3)}$

$\footnotesize\implies n_{(MgCO_3)} = \dfrac{8.4}{36 +48}$

$\footnotesize\implies n_{(MgCO_3)} = \dfrac{8.4}{84}$

$\footnotesize\implies \bf \red{ n_{(MgCO_3)} = 0.1 \: mol}$

Write the chemical equation and balance it:

$\footnotesize\sf MgCO_{3}\overset{\Delta}{\longrightarrow}\:MgO+ CO_{2}$

From 1 mole of MgCO₃ produce moles of CO₂ = 1

Therefore,0.1 mole of MgCO₃ produce moles of CO₂ = 0.1

Relation of mole with volume of gas at STP:

1 mole of CO₂ gas at STP has volume = 22.4 L

Therefore, 0.1 mole of of CO₂ gas at STP has volume = 22.4 × 0.1 = 2.24 L

Hence, Carbon dioxide released into the atmosphere at S.T.P. will be 2.24 L.

Answer 2

Answer:

Explanation:

see the attached attached