0.5 mol CaCO_3 solid decompose in 500 mL heated in closed vessel at 400 K reaction CaCO_(3(s)) hArr CaO_((s)) + CO_(2(g)) equilibrium constant of K_c = 0.9 "mol L"^(-1). Calculate mol of CO_2 at equilibrium how much percentage of reaction completed ?

0.5 mol CaCO_3 solid decompose in 500 mL heated in closed vessel at 40

1.	0.5 mol CaCO_3 solid decompose in 500 mL heated in closed vessel at 400 K reaction CaCO_(3(s)) hArr CaO_((s)) + CO_(2(g)) equilibrium constant of K_c = 0.9 "mol L"^(-1). Calculate mol of CO_2 at equilibrium how much percentage of reaction completed ?
Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`[CO_2]= 0.45 "<a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a> L"^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)`, 90% decomposition <br/>`[CO_2]`= 0.3 mol and <a href="https://interviewquestions.tuteehub.com/tag/60-328817" style="font-weight:bold;" target="_blank" title="Click to know more about 60">60</a>% <a href="https://interviewquestions.tuteehub.com/tag/reaction-22747" style="font-weight:bold;" target="_blank" title="Click to know more about REACTION">REACTION</a> if `CaCO_3` and CaO used in calculation</body></html>