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Quantitative chemistry (1 Viewer)
- Thread starter hs17
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Mol of H2S =
 = 0.469 mol)
Mol of SO2 =
 = 0.312 mol)
This means that since the ratio should be 2:1, andn since 0.469/2 = 0.235 (Rounding), the H2S is the limiting reagent.
This means remaining SO2 = 0.312 - 0.469/2 = 0.077
As such, the remaining mass is 0.077*(16*2+32.065) = 4.93 grams of SO2(3 S.F)
Apologies for all the edits - I am a bit dumb and didn't realise the SO2 wasn't 2SO2
Mol of SO2 =
This means that since the ratio should be 2:1, andn since 0.469/2 = 0.235 (Rounding), the H2S is the limiting reagent.
This means remaining SO2 = 0.312 - 0.469/2 = 0.077
As such, the remaining mass is 0.077*(16*2+32.065) = 4.93 grams of SO2(3 S.F)
Apologies for all the edits - I am a bit dumb and didn't realise the SO2 wasn't 2SO2
Last edited:
Velocifire
Critical Hit
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- 2021
Great explanation, thanks.