(RAM H = 1.0, C = 12.0 and 1 mol of gas occupies 24.5 litres at LTP)
Which is the limiting reagent when 100.0g of hexane C6H14(l) is reacted with 500.0 litres of oxygen gas at LTP, according to the reaction: 2 C6H14(l) + 19 O2(g) > 12 CO2(g) + 14 H2O(l)
Jawaban Terkonfirmasi
N C6H14 = 100/86 = 1,163 mol
n O2 = 500/24,5 = 20,41 mol
2 C6H14 + 19 O2 —-> 12 CO2 + 14 H2O
1,163 20,41 - – n mula
1,163 11,0485 6,978 8,141 n reaksi
– 9,3615 n akhir
n sisa = O2, jadi C6H14 adalah pereaksi pembatasnya