Quantitative chemistry (1 Viewer)

hs17

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The answers in the textbook are a bit off with the rounding...so could someone please do part b) ?
 

Pedro123

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Mol of H2S =

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
 
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