Question

in the combustion of hydrogen gas, hydrogen reacts with oxygen from the air to form water vapor, if you burn 50.0g of hydrogen and produce 447g of water how much oxygen is reacted

Answer 1

Approximately 12.375 moles of oxygen are reacted in the combustion of 50.0g of hydrogen gas.

Step-by-step explanation:

The balanced chemical equation for the combustion of hydrogen gas is:

2H₂(g) + O₂(g) → 2H₂O(l)

From the equation, we can see that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water. We can use stoichiometry to find the amount of oxygen reacted.

Given that 50.0g of hydrogen reacts to produce 447g of water, let's first convert the mass of hydrogen to moles:

• Molar mass of hydrogen (H₂): 2.02 g/mol
• Moles of hydrogen: 50.0 g ÷ 2.02 g/mol = 24.75 mol

Now, we can use the stoichiometric ratio between hydrogen and oxygen:

• 1 mole of hydrogen reacts with 0.5 moles of oxygen
• 24.75 moles of hydrogen will react with 0.5 × 24.75 = 12.375 moles of oxygen

Therefore, approximately 12.375 moles of oxygen are reacted.

Answer 2

2H₂ ₍g₎ + O₂ ₍g₎ --------> 2H₂O ₍l₎

If mass of water = 447 g
And mol =
`(MASS)/(MOLAR  MASS)`

then mol of H₂O =
`(447 g)/(((16)+(1 * 2) g / mol)`
= 24.83 mol

Mole ratio of H₂O : O₂ ≡ 2 : 1

Since mole of H₂O = 24.83 mol
then mole of O₂ = 24.83 mol ÷ 2
= 12.42 mol

Thus Mass of O₂ = 12.42 mol * (16 * 2) g / mol
= 397.28 g