Table salt, NaCl(s), and sugar, C12H22O11(s), are accidentally mixed. A 6.00-g sample is burned, and 2.80...

Table salt, NaCl(s), and sugar, C12H22O11(s), are accidentally mixed. A 6.00-g sample is burned, and 2.80 g of CO2(g) is produced. What was the mass percentage of the table salt in the mixture?

Expert Solution

Solution :-

combustion of sugar balanced equation

C12H22O11 + 12O2 ----- > 12CO2 + 11H2O

lets calculate the moles of the CO2

2.80 g CO2 * 1 mol / 44.01 g = 0.06362 mol CO2

now using the moles of CO2 lets calculate the moles of sugar

0.06362 mol CO2 * 1 mol C12H22O11 / 12 mol CO2 = 0.05302 mol C12H22O11

mass of sugar = moles * molar mass

=0.05302 mol * 342.3 g per mol

= 1.81 g sugar

mass of table salt = 6.00 g - 1.815 g = 4.185 g

now lets calculate the percent of the table salt

% table salt = (mass of table salt / mass of mixture)*100%

= (4.185 g / 6.00 g)*100%

= 69.75 % table salt

queen_honey_blossom answered 2 years ago

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