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mariarad [96]
2 years ago
10

A 0. 1220 g vitamin c tablet was dissolved in acid. This required 11. 50 ml of 0. 01740 m kio3 to reach the endpoint. Calculate

percent by weight of ascorbic acid in the tablet
Chemistry
1 answer:
kodGreya [7K]2 years ago
8 0

86.66 %  by weight of ascorbic acid in the tablet, when 0. 1220 g vitamin c tablet was dissolved in acid. This required 11. 50 ml of 0. 01740 m KIO_3 to reach the endpoint.

<h3>What is a balanced chemical equation?</h3>

A balanced chemical reaction is an equation that has equal numbers of each type of atom on both sides of the arrow.

3C_6H_80_6 +3I_2 +KIO_3 +5KI +6H^+ → 3C_6H_60_6 +6I^- +3I_2+6K^+ +3H_2O

So, the balanced chemical equation for the reaction will be:

3C_6H_8O_6 +KIO_3 → 3C_6H_6O_6 +KI + 3H_2O

That means, 3 mol of of ascorbic acid reacts with 1 mol of KIO_3

Moles of KIO_3 available is 0.0174 X 0.0115 =0.0002001 mol

Moles of ascorbic acid to be yielded should be 3 X 0.0002001 =0.0006003

So, percent mass of ascorbic acid in the tablet will be:

(0.1057 / 0.1220) X 100 %

=86.66 %

Hence, the third option is the correct answer.

Learn more about balanced chemical equations here:

brainly.com/question/26750249

#SPJ4

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Co(H_2O)_6^{3+}(aq)\rightleftharpoons H^+(aq)+Co(H_2O)_5(OH)^{2+}(aq)

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The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

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HC_2H_3O_2(aq)\rightleftharpoons H^+(aq)+C_2H_3O_2^-(aq)

The equilibrium expression of HC_2H_3O_2 will be:

K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}

(B) The dissociation reaction of Co(H_2O)_6^{3+} will be:

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The equilibrium expression of Co(H_2O)_6^{3+} will be:

K_a=\frac{[H^+][Co(H_2O)_5(OH)^{2+}]}{[Co(H_2O)_6^{3+}]}

(C) The dissociation reaction of CH_3NH_3^+ will be:

CH_3NH_3^+(aq)\rightleftharpoons H^+(aq)+CH_3NH_2(aq)

The equilibrium expression of CH_3NH_3^+ will be:

K_a=\frac{[H^+][CH_3NH_2]}{[CH_3NH_3^+]}

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