Number of moles of oxygen required for the combustion of 48 g of methane ________.

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Solution

Balanced chemical equation for combustion of methane as follows.
$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$
For 1 mole of combustion of methane 2 mole of oxygen are required.
48-gram methane $= \frac{48}{16} = 3$ moles of methane.(Molecular weight of methane is 16)
So, no. of moles of oxygen required will be 6.

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Number of moles of oxygen required for the combustion of 48 g of methane ________.

Balanced chemical equation for combustion of methane as follows.CH4+2O2→CO2+2H2OFor 1 mole of combustion of methane 2 mole of oxygen are required.48-gram methane =4816=3 moles of methane.(Molecular weight of methane is 16)So, no. of moles of oxygen required will be 6.