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Juliette [100K]
3 years ago
7

7.8g of dioxide would be produced from how many moles of potassium chlorate

Chemistry
1 answer:
Ghella [55]3 years ago
6 0
2KClO₃ ⇒ 2KCl + 3O₂
2mol         :            3mol
245,1g      :           95,94g
x               :           7,8g

x = 7,8g*245,1g / 95,94g = 19,93g

1mol of KClO₃ --------- is --------- 122,55g
x mol               --------- is --------- 19,93g

x = 19,93g*1mol / 122,55g = 0,158mol
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What is the difference between liquid nitrogen and nitrogen gas​
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A student is given a 2.002 g sample of unknown acid and is told that it might be butanoic acid, a monoprotic acid (HC4H7O2, equa
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<u>Answer:</u> The identity of the unknown acid is butanoic acid or ascorbic acid.

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}

Molarity of NaOH solution = 0.570 M

Volume of solution = 39.55 mL

Putting values in above equation, we get:

0.570M=\frac{\text{Moles of NaOH}\times 1000}{39.55}\\\\\text{Moles of NaOH}=\frac{0.570\times 39.55}{1000}=0.0225mol

The chemical equation for the reaction of NaOH and monoprotic acid follows:

NaOH+HX\rightarrow NaX+H_2O

By Stoichiometry of the reaction:

1 mole of NaOH reacts with 1 mole of HX

So, moles of monoprotic acid = 0.0225 moles

The chemical equation for the reaction of NaOH and diprotic acid follows:

2NaOH+H_2X\rightarrow 2NaX+2H_2O

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of diprotic acid

So, moles of diprotic acid = \frac{0.0225}{2}=0.01125moles

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

  • <u>For butanoic acid:</u>

Mass of butanoic acid = 2.002 g

Molar mass of butanoic acid = 88 g/mol

Putting values in above equation, we get:

\text{Moles of butanoic acid}=\frac{2.002g}{88g/mol}=0.02275mol

  • <u>For L-tartaric acid:</u>

Mass of L-tartaric acid = 2.002 g

Molar mass of L-tartaric acid = 150 g/mol

Putting values in above equation, we get:

\text{Moles of L-tartaric acid}=\frac{2.002g}{150g/mol}=0.0133mol

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Mass of ascorbic acid = 2.002 g

Molar mass of ascorbic acid = 176 g/mol

Putting values in above equation, we get:

\text{Moles of ascorbic acid}=\frac{2.002g}{176g/mol}=0.01137mol

As, the number of moles of butanoic acid and ascorbic acid is equal to the number of moles of acid getting neutralized.

Hence, the identity of the unknown acid is butanoic acid or ascorbic acid.

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