If you burn 45.2g of hydrogen and produce 404g of water, how much oxygen reacted

45.2÷2= 22.6 moles of hydrogen 2H2 + O2 = 2H2O 404g÷ 18= 22.4 moles of water 2 moles of H2 reacts with 1 mole of O2 So 22.6 will react with 11.3 moles of O2 11.3×32= 361.6 grams.

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Ondinne Ondinne · Answer 2 · 2019-10-16T18:03:30+02:00

Answer:

[tex]359gO_{2}[/tex]

Explanation:

1. Write the balanced equation:

[tex]_{2}H_{2}+O_{2}=_{2}H_{2}O[/tex]

2. Find the quantity of [tex]H_{2}O[/tex] produced by 45.2g of hydrogen:

[tex]45.2gH_{2}*\frac{1molH_{2}}{2.016gH_{2}}*\frac{2molesH_{2}O}{2molesH_{2}}*\frac{18gH_{2}O}{1molH_{2}O}= 404gH_{2}O[/tex]

So, 45.2g of [tex]H_{2}[/tex] produce 404g of [tex]H_{2}O[/tex]

3. Find the quantity of [tex]O_{2}[/tex] that reacted with 45.2g of [tex]H_{2}[/tex]:

[tex]404gH_{2}O*\frac{1molH_{2}O}{18gH_{2}O}*\frac{2molesO_{2}}{2molesH_{2}O}*\frac{16gO_{2}}{1molO_{2}}=359gO_{2}[/tex]