What mass of oxygen is needed for the complete combustion of 2.60×10−3 of methane?

Express your answer with the appropriate units.

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http://www.jiskha.com/science/chemistry/stoichiometry.html

To find the mass of oxygen needed for the complete combustion of methane, we first need to balance the chemical equation for the combustion of methane.

The balanced equation is:
CH4 + 2O2 -> CO2 + 2H2O

From the balanced equation, we can see that for every 1 mole of methane (CH4) reacting, we need 2 moles of oxygen (O2) to completely combust.

The molar mass of methane (CH4) is 16.04 g/mol.

Using the given amount of methane, 2.60×10^-3 mol, we can calculate the moles of oxygen needed as follows:

2.60×10^-3 mol CH4 * (2 mol O2 / 1 mol CH4) = 5.20×10^-3 mol O2

The molar mass of oxygen (O2) is 32.00 g/mol.

Now, to find the mass of oxygen needed, we can multiply the moles of oxygen by its molar mass:

5.20×10^-3 mol O2 * 32.00 g/mol = 0.1664 g

Therefore, the mass of oxygen needed for the complete combustion of 2.60×10−3 mol of methane is 0.1664 grams.