asked 209k views
5 votes
10g of hyrogen react with excess of oxygen gas according to the equation:

2H2+O2---->2H2O
what is the reacted volume of oxygen gas at STP and the mass of the produced water vapour in this reaction ?

1 Answer

7 votes

Answer:

Volume of O₂ = 56 dm³

mass of water vapors (H₂O) = 90 g

Step-by-step explanation:

Data Given:

mass of Oxygen = 10 g

Volume of Oxygen = ?

mass of the water vapor = ?

Reaction Given:

2H₂+O₂---->2H₂O

Solution:

First we have to look at the reaction for the information required

2H₂ + O₂ -------> 2H₂O

2 mol 1mol 2 mol

now convert moles to grams

molar mass of H₂ = 2(1) = 2 g/mol

molar mass of O₂ = 2(16) = 32 g/mol

molar mass of H₂0 = 2(1) + 16 = 18 g/mol

So the masses will be

2H₂ + O₂ -------> 2H₂O

2 mol (2 g/mol) 1mol (32 g/mol) 2 mol (18 g/mol)

4 g 32 g 36 g

So now we know that

4 g of hydrogen combine with 32 g of Oxygen and give 36 g of water vapors.

By using above information

First we find the volume of Oxygen:

For this first we find mass and then moles of Oxygen

As we know

if 4 g of hydrogen combine with 32 g of Oxygen then how much oxygen will react with 10 g of hydrogen

Apply unity formula

4 g of hydrogen H₂ ≅ 32 g of Oxygen O₂

10 g of hydrogen H₂ ≅ X g of Oxygen O₂

by doing Cross multiplication

g of Oxygen O₂ = 32 g x 10 g / 4 g

g of Oxygen O₂ = 80 g

So,

mass of oxygen = 80 g

now find moles of oxygen

formula used:

no. of moles = mass in grams/ molar mass . . . . . . (1)

Put values in above equation 1

no. of moles = 80 g / 32 g/mol

no. of moles = 2.5

Now to find volume of oxygen

Formula used

Volume of O₂ = no. of moles x molar volume (22.4 dm³/ mol) . . . . . . (2)

Put values in equation 2

Volume of O₂ = 2.5 moles x 22.4 dm³/mol

Volume of O₂ = 56 dm³

______________________

Now to find mass of water vapors

As we now

if 4 g of hydrogen produce 36 g of water vapors then how much water vapor will produce from 10 g of hydrogen

Apply unity formula

4 g of hydrogen H₂ ≅ 36 g of water vapors (H₂O)

10 g of hydrogen H₂ ≅ X g of water vapors (H₂O)

by doing Cross multiplication

g of water vapors (H₂O) = 36 g x 10 g / 4 g

g of water vapors (H₂O) = 90 g

So,

mass of water vapors (H₂O) = 90 g

answered
User MatthieuGD
by
8.0k points
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