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21 March, 16:13

Table salt, nacl (s), and sugar, c12h22o11 (s), are accidentally mixed. a 3.50-g sample is burned, and 2.40 g of co2 (g) is produced. what was the mass percentage of the table salt in the mixture?

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  1. 21 March, 16:24
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    First we will write the formula for the combustion of the sugar.

    C₁₂H₂₂O₁₁ + 12O₂ → 12CO₂ + 11H₂O

    We have produced 2.40 g of CO₂. We will convert this to moles of CO₂ and then convert that value to moles of sugar.

    2.40 g / 44.0 g/mol = 0.0545 moles CO₂ x 1 mole C₁₂H₂₂O₁₁/12 mole CO₂ = 0.00455 moles of C₁₂H₂₂O₁₁

    We can now convert the moles of sugar to a mass of sugar.

    0.00455 moles C₁₂H₂₂O₁₁ x 342 g/mol = 1.56 g C₁₂H₂₂O₁₁

    We want to determine the mass percentage of salt in the 3.5 g mixture of sugar and salt.

    3.5 g mixture - 1.56 g sugar = 1.94 g salt

    (1.94 g salt / 3.5g mixture) x 100% = 55%

    Therefore, the mass percentage of table salt in the mixture is 55%.
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