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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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You are given sodium acetate, 1m hcl, nahco3 and na2co3. Determine which of these four you would need and then show calculations
valentina_108 [34]

solution:

the given compounds are sodium acetate, 1M HCL,NaHCO₃ and Na2CO₃

pH of the buffer solution is 4.7

the value of pKa of sodium bicarbonate is 6.37

the value of pKa of acetic acid is 4.7

calculate concentration of acetic acid by using the following forumula

pH=pKa+lag[salt]/[acid]

substitute the pH and Pka values in the formula.

4.7=4.7+log[salt]/[acid]

log[salt]/[acid]=0

thus, the concentration ratio of the salt and acid should be equal to each other.

Thus, concentration of sodium acetate is 0.05M

Concentration of sodium acetate= concentration of acid

= 0.05M

Volume of the buffer solution is 100mL

The buffer solution can be prepared as 0.05M of 50mL sodium acetate will react with 0.05M of 50mL of 0.05M of HCL.

The chemical equation for neutralization of the weak base with strong can be represented as show as

CH₃COONa+HCL-->CH₃COOH+NaCL


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3 years ago
What type of compound is almost always found as a solid?
Evgesh-ka [11]
Hi , your answer is Ionic.
4 0
3 years ago
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What property of water enables ice wedging to break rock?
seropon [69]

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A student has a salt water solution. When the student adds more salt to the solution it completely dissolves. What type of solut
AlladinOne [14]

Answer:

<h2>Saturated</h2>

Explanation:

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2 years ago
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1.12g H2 is allowed to react with 9.60 g N2, producing 1.23 g NH3.
andriy [413]

Answer:

A. m_{NH_3}^{theo} =1.50gNH_3

B. Y=82.2\%

Explanation:

Hello!

In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:

N_2+3H_2\rightarrow 2NH_3

Thus we proceed as follows:

A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:

n_{NH_3}^{by \ H_2}=1.12gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.370molNH_3\\\\  n_{NH_3}^{by \ N_2}=1.23gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.0878molNH_3

Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:

m_{NH_3}^{theo}=0.0878mol*\frac{17.04gNH_3}{1molNH_3} =1.50gNH_3

B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:

Y=\frac{1.23gNH_3}{1.50gNH_3} *100\%\\\\Y=82.2\%

Best regards!

5 0
3 years ago
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