Question

20 gram of magnesium carbonate sample decomposes on heating to give ac co2 and 8 gram of mg oh what will be the percentage purity of magnesium carbonate in the sample

Answer 1

Answer:

Number of Moles -

No of moles = given mass of substance/ molar mass of substance

-

MgCO_{3} \rightarrow MgO(s)+CO2(g)

supposed meals of = \frac{mass \:\:in\:\: grams}{molar \:\:mass}

= \frac{20}{84} = 0.238 \:\:mal

By the stoichoimetry of the reaction alone, we know that 1 mal of the MgCO_{3} gives 1 mal of MgO moles of MgO obtained

= \frac{8}{40} = 0.2 moles of MgO that should have been obtained in a pure sample = 0.238

\therefore moles of MgCO_{3} in the sample

= moles of MgO in the sample

= 0.2