How many moles of oxygen will be needed to completely oxidize 64g CH4?

Answer 1

Molar mass:

• O2 32 g/mol
• CH4 16 g/mol

Reaction:

• CH4 + 2 O2 --> CO2 + 2H2O

Calculus

• 1 mol CH4 with 2 moles O2
• Given: 24 g CH4 equals 24/16 = 1.5 mol CH4
• this will need 2 * 1.5 moles O2 = 3.0 mol O2
• 3.0 mol O2 equals 3.0 * 32 = 96 g O2

Answer 2

The equation for the combustion of methane is CH4 + 2 O2 -> CO2 + 2 H2O. This means that for every mole of methane, two moles of oxygen are needed. In .009 g of methane, there are .00561 moles (molar mass of methane: 16.0425 g/mol). So there are .01122 moles of oxygen, meaning that with a molar mass of 31.9988 g/mol, there are .18 grams of O2 needed.

Answer 3

what is the mass in grams of oxygen, is needed to complete combustion of 6 L of methane?

Answer 4

take the balanced equation CH4+2O2---->CO2+2H2O.16g need 32g of O2.So 74.5g need 149g of O2

Answer 5

2C2H6+7O2----->4CO2+6H2O

74g of C2H6= 2.4667 mol of C2H6

from the ratio(2/7), 8.633 mol of O2 are needed which is 276.256g (8.633mol×32g(molar mass))

Answer 6

7,98 moles of oxygen are needed to completely oxidize 64g CH4.

Answer 7

The equation for complete combustion of CH4 is CH4 + 2O2 ==> CO2 + 2H2OSo, each 1 mole of CH4 requires 2 moles of O2. Therefore 3.0 moles of CH4 requires 6.0 moles O2.

Answer 8

6 moles