How many grams of oxygen gas is essentially required for complete combustion of 3 moles of butane gas?
(1) 624 g
(2) 312 g
(3) 128 g
(4) 64 g
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How many grams of oxygen gas is essentially required for complete combustion of 3 moles of butane gas?
(1) 624 g
(2) 312 g
(3) 128 g
(4) 64 g
How many grams of oxygen gas is essentially required for complete combustion of 3 moles of butane gas?
(1) 624 g
(2) 312 g
(3) 128 g
(4) 64 g
(1) 624 g
2 C4 H10 + 13O2 → 8 CO2 + 10 H2 O
2 moles required 13 mole O2
1 mole required 13/2 mole O2
3 mole required 13/2 x 3 = 39/2 = 19.5 mole
Wt of O2 required = 19.5 x 32 = 624 g