Chem A-Level PPQs
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(can’t say red litmus dipped in solution)
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Tests for Grp2, Ammonium and Sulfates
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Tests for hydroxides, halides and carbonates
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1 mol C8H18 needs 12.5 mol of O2
So, 0.0010 mol C8H18 needs: 0.0010×12.5 = 0.0125mol O2
There is 0.0200mol O2 available, meaning O2 is in excess, and C8H18 is limiting
CO2 produced = 0.0010 × 8 = 0.0080mol
O2 used = 0.0125mol
O2 remaining = 0.0200-0.0125 = 0.0075mol
Total gas = CO2 + leftover O2 = 0.0080 + 0.0075 = 0.0155mol
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