How many moles of sulphur dioxide are produced when 72.0 grams of water is produced by the process: 2H2S + 3O2 --> 2H2O + 2SO

Question

Yanka [14]

3 years ago

15

How many moles of sulphur dioxide are produced when 72.0 grams of water is produced by the process: 2H2S + 3O2 --> 2H2O + 2SO

2

Chemistry

1 answer:

Lelu [443] · Answer 1 · 2022-05-04T05:38:40+03:00

Molar mass of sulphur dioxide ( SO2) = 64.066 g/mol
Molar mass of water ( H2O ) = 18.0 g/mol

process:

2 H2S + 3 O2 --> 2 H2O + 2 SO2
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2 * 18 g H2O --------------> 2* 64.066 g SO2
72.0 g H2O ----------------> mass of SO2

mass of SO2 = 72.0 * 2 * 64.066 / 2 * 18

mass of SO2 = 9225.504 / 36

mass of SO2 = 256.264 g

1 mole --------------> 64.066 g
mole SO2 ----------> 256.264

mole SO2 = 256.264 / 64.066

= 4 moles of SO2

hope this helps!.

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