Answer:
Therefore , 37.27g of COâ‚‚ unreacted or remained because KOâ‚‚ is limiting
Explanation:
The balanced chemical equation is
[tex]4K0_2(s) + 2CO_2(g) \rightarrow 2K_2CO_3(s) + 3O_2(g[/tex]
we calculate the moles of KOâ‚‚ and COâ‚‚
[tex]Moles=\frac{Mass}{Molecular\:mass}[/tex]
Moles of KOâ‚‚ = 25.0/71.1 = 0.3516 moles
Moles of COâ‚‚ = 45.0/44 = 1.0227 moles
[tex]4K0_2(s) + 2CO_2(g) \rightarrow 2K_2CO_3(s) + 3O_2(g[/tex]
1 mole of KOâ‚‚ requires 1/2 mole of COâ‚‚
0.3516 moles  of KO₂ requires [tex]=\frac{1}{2} \times0.3516=0.1758[/tex]
∴ no of mole of CO₂ required to complete reaction is 0.1758 moles
Avalaible no of mole of COâ‚‚ is 1.0227 moles
∴ The exces no of mole of CO₂ is 1.0227 - 0.1758
= 0.8469 moles of COâ‚‚
∴ mass of CO₂ that did not react is
no of mole x molar mass
=0.8469 x 44.01
= 37.27g of COâ‚‚
Therefore , 37.27g of COâ‚‚ unreacted or remained because KOâ‚‚ is limiting