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3 years ago

If 3 moles of H2 react with two moles of o2 what is limiting reactant

Chemistry

1 answer:

Dominik [7]3 years ago

8 0

H₂ is the limiting reactant.

<u>Explanation:</u>

H₂ reacts with O₂

The reaction would be

2H₂ + O₂ → 2H₂O

According to the balanced equation, 2 moles of H₂ reacts with 1 mole of O₂ to form 2 moles of H₂O.

The ratio of usage of H₂ and O₂ is 2 : 1 respectively

If 3 moles of H₂ and 2 moles of O₂ are present then:

3 moles of H₂ would require 1.5 moles of O₂ ( 2 : 1 of H₂ and O₂ )

Out of 2 moles of O₂, 1.5 moles would be used and 0.5 mole would be in excess.

Therefore, H₂ is the limiting reactant as the number of moles of H₂ are not enough to use all the O₂.

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when atoms react, they often lose, gain, or share electrons to form a more stable version of themselves. for example, alkali met

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<h3></h3><h3>What is periodic table?</h3>

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1 year ago

Read 2 more answers

An unknown amount of acid can often be determined by adding an excess of base and then back-titrating the excess. A 0.3471−g sam

MAXImum [283]

Explanation:

The given data is as follows.

Mass of mixture = 0.3471 g

As the mixture contains oxalic acid and benzoic acid. So, oxalic acid will have two protons and benzoic acid has one proton.

This means oxalic acid will react with 2 moles of NaOH and benzoic acid will react with 1 mole of NaOH.

Hence, moles of NaOH in 97 ml = $\frac{0.1090 \times 97}{1000}$

= $10.573 \times 10^{-3} mol$

Moles of HCl in 21.00 ml = $\frac{0.2060 \times 21}{1000}$

= $4.326 \times 10^{-3}$ mol

Therefore, total moles of NaOH that reacted are as follows.

$10.573 \times 10^{-3} mol$ - $4.326 \times 10^{-3} mol$

= $6.247 \times 10^{-3}$ mol

So, total 3 mole of NaOH will react with 1 mole of mixture. Therefore, number of moles of NaOH reacted with benzoic acid is as follows.

$\frac{6.247 \times 10^{-3}}{3}$

= $2.082 \times 10^{-3}$ mol

Since, molar mass of NaOH is 40 g/mol. Therefore, calculate the mass of NaOH as follows.

$2.082 \times 10^{-3} mol \times 40 g/mol$

= $83.293 \times 10^{-3}$ g

= 0.0832 g

Whereas molar mass of benzoic acid is 122 g/mol.

Therefore, 40 g NaOH = 122 g benzoic acid

So, 0.0832 g NaOH = $\frac{122 g}{40 g} \times 0.0832 g$

= 0.253 g

Hence, calculate the % mass of benzoic acid as follows.

$\frac{0.253 g}{0.3471 g} \times 100$

= 73.10%

Thus, we can conclude that mass % of benzoic acid is 73.10%.

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