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Irina18 [472]
3 years ago
8

4 moles of hydrogen react with 5 moles of oxygen to form water, identify the excess reagent?

Chemistry
1 answer:
Fynjy0 [20]3 years ago
5 0

Answer:

Excess Reagent = oxygen

Explanation:

Limiting reagent: The substance that is totally consumed when the reaction is completed.

Excess reagent: The substance left after the limiting reagent is consumed completely

The balanced chemical equation for formation of water is as follow:

2H_{2}+O_{2}\rightarrow 2H_{2}O

This means when 2 moles of hydrogen reacts with 1 mole of oxygen, 2 moles of water is produced.

Hence the ratio in which hydrogen and oxygen gas reacts is 2:1

Now if 2 mole hydrogen require 1 mole of oxygen ,then 4 mole hydrogen need 2 mole of oxygen.

H_{2}:O_{2}=2:1

or

H_{2}:O_{2}=4:2

Here 5 mole of oxygen is reacting but only 2 mole is required .

Oxygen is in excess.

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How many moles of na2co3 are necessary to reach stoichiometric quantities with cacl2
lbvjy [14]

0.0102 moles Na₂CO₃ = 1.08g of Na₂CO₃ is necessary  to reach stoichiometric quantities with cacl2.

<h3>Explanation:</h3>

Based on the reaction

CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃

1 mole of CaCl₂ reacts per mole of Na₂CO₃

we have to calculate how many moles of CaCl2•2H2O are present in 1.50 g

  • We must calculate the moles of CaCl2•2H2O using its molar mass (147.0146g/mol) in order to answer this issue.
  • These moles, which are equal to moles of CaCl2 and moles of Na2CO3, are required to obtain stoichiometric amounts.
  • Then, we must use the molar mass of Na2CO3 (105.99g/mol) to determine the mass:

<h3>Moles CaCl₂.2H₂O:</h3>

1.50g * (1mol / 147.0146g) = 0.0102 moles CaCl₂.2H₂O = 0.0102moles CaCl₂

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0.0102 moles Na₂CO₃

Mass Na₂CO₃:

0.0102 moles * (105.99g / mol) = 1.08g of Na₂CO₃ are present

Therefore, we can conclude that 0.0102 moles Na₂CO₃  is necessary.to reach stoichiometric quantities with cacl2.

To learn more about stoichiometric quantities visit:

<h3>brainly.com/question/28174111</h3>

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