the balanced equation is as follows
2 C4H10 + 13 O2 -----------------------> 8 CO2 + 10 H2O
no of moles = mass / molar mass
for C4H10 = 26.0 g / 58.12 g/mol => 0.447 mol
for O2 = 59.0 g / 32 g/mol => 1.84375 mol
=> 13 mol O2 -----------------> 2 mol C4H10
1.84375 mol -------------------->?
=> 1.84375 * 2 / 13
=> 0.2836 mol of C4H10 required
but excess moles of C4H10 is present
excess moles unreacted => 0.447 - 0.2836 => 0.1637 moles
mass of butane left over is => 0.1637 * 58.12 => 9.514g
answer => 9.5 g
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