Question 1

Pentane (c5h12) undergoes combustion with excess oxygen to produce water and carbon dioxide. how many liters of water are produced when 50.5 grams of pentane combustion with excess oxygen at stp?

Question 2

$50.5 g C_(5){H}_(12) Molar mass C_(5)H_(12) =5*12+12*1=72 g/mol 50.5 g* (1 mol/72 g)=0.706 mol C_(5)H_(12) C_(5)H_(12) + 8O_(2) ---\ \textgreater \ 5CO_(2) +6 H_(2)O 1 mol C_(5)H_(12) ----- 6 mol H_(2)O 0.706 mol C_(5)H_(12) --- x mol H_(2)O x= (0.706*6)/1= 4.24 mol H_(2)O Under STP water is liquid. M(H_(2)O})=2*1.0+16.0=18.0 (g)/(mol) 4.24 mol H_(2)O * (18.0 g)/(1 mol) = 76.3 g H_(2)O Density H_(2)O = 1 (g)/(ml) 76.3 g /(1 g/ml) = 76.3 ml of water Answer 76.3 ml of the water.$

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