A sample containing an inert substance and sodium carbonate is heated. The mass

Question

A sample containing an inert substance and sodium carbonate is heated. The mass of the sample before heating was 3.327g and after heating the mass was 2.432g. What is the percent composition of sodium carbonate in the original mixture?

Explanation / Answer

Na2CO3 -> Na2O + CO2 amount of CO2 released = (3.3327-2.432)g = 0.895 gm. Moles of CO2 = 0.895/44 = 0.02034 moles. By stoicheometry Moles of Na2O produced = 0.02034 moles = 0.02034*62 = 1.261 gm Na2O. Hence, inert substance present initially = 2.432-1.261 = 1.171 gm Hence, In initial mixture Na2CO3 will be = 3.327 - 1.171 = 2.156 gm. So, % of Na2CO3 in initial mixture = 2.156/3.327 * 100 = 64.8 %. this is final answer.

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