Sugar, C 12 H 22 O 11 undergoes decomposition of heating to produce CO 2 and wat

Question

Sugar, C12H22O11 undergoes decomposition of heating to produce CO2 and water as per the reaction.

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

As per the stoichiometry of the reaction,

1 mole C12H22O11 = 12 moles CO2.

Molar mass of CO2 = (1*12.01 + 2*15.9994) g/mol = 44.0088 g/mol.

The combustion reaction produced 2.10 g CO2; therefore, mole(s) of CO2 produced = (2.10 g)*(1 mole/44.0088 g) = 0.04772 mole.

Moles of C12H22O11 which produces 0.0477 mole CO2 is (0.04772 mole CO2)*(1 mole C12H22O11/12 mole CO2) = 0.003977 mole.

Molar mass of C12H22O11 = (12*12.01 + 22*1.008 + 11*15.9994) g/mol = 342.2894 g/mol.

Mass of C12H22O11 corresponding to 0.003977 mole = (0.003977 mole)*(342.2894 g/mol) = 1.3613 g.

Mass of table salt in the mixture = (3.50 – 1.3613) g = 2.1387 g.

Mass percentage of table salt in the mixture = (2.1387 g)/(3.50 g)*100 = 61.1057 61.11% (ans).

Explanation / Answer

Table salt, NaCI(s), and sugar, C2Hz2011(s), are accidentally mixed. A 3.50-g sample is burned, and 2.10 g of COg) is produced. What was the mass percentage of the table salt in the mixture? Number 40.86 % NaCl

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