Table salt, NaCl(S), and sugar, C12H22O11 (s), are accidentally mixed. A 4.50-g

Question

Table salt, NaCl(S), and sugar, C12H22O11 (s), are accidentally mixed. A 4.50-g sample is burned and 2.90 g of CO2 (g) is produced. What was the mass percentage of the table salt in the mixture?


Hint"Only the sugar burns , start by writing chemical equation for combustion of C12H22O11"

Explanation / Answer

Here the balanced reaction would be C12H22O11 + 12 O2 =======> 12 CO2 + 11 H2O So the molar ratio is 1 C12H22O11 : 12 CO2 2.90 g CO2 * (1 mol CO2 / 44 g CO2)*(1 molC12H22O11/ 12 mol CO2)*( 342.3 g C12H22O11/1 mol C12H22O11) 1.88 g C12H22O11 So mass of table salt = 4.50-1.88 = 2.62 g So Mass % = mass of the substance/ total mass * 100 = (2.62/4.50)*100 = 58.22 % Table salt

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