Question

When 14.51 grams of methane (CH4) are burned (see equation below), how many moles of oxygen gas (O2) will be used? Please round your answer to two digits after the decimal point and don't forget units and substance!

CH4 + 2 O2 --> CO2 + 2 H2O

Answer 1

When 14.51 grams of methane are burned, the number of moles of oxygen gas used can be determined by using the balanced chemical equation:

CH4 + 2 O2 -> CO2 + 2 H2O

Since the coefficients in the equation represent the mole ratios of the reactants and products, we can use the number of moles of methane as the basis for calculating the number of moles of oxygen.

First, we'll calculate the number of moles of methane:

moles of CH4 = 14.51 g / 16.04 g/mol = 0.905 mol

Next, we'll use the mole ratio of CH4 to O2 to determine the number of moles of O2 required for the reaction:

moles of O2 = 2 * moles of CH4 = 2 * 0.905 mol = 1.81 mol

So, the amount of oxygen used is 1.81 moles.

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