Question

When 19.04 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

The number of moles of oxygen gas used when 19.04 grams of methane is burned is 2.38 mol O2.

Step-by-step explanation:

To answer this question, we need to use the balanced chemical equation provided and the given mass of methane to determine the number of moles of oxygen gas used.

First, we need to calculate the number of moles of methane present in 19.04 grams of CH4:

Number of moles of CH4 = mass of CH4 / molar mass of CH4

= 19.04 g / 16.04 g/mol

= 1.19 mol

Next, we can use the stoichiometric coefficients from the balanced equation to determine the number of moles of oxygen gas required to react with the 1.19 mol of CH4:

1 mol CH4 + 2 mol O2 → 1 mol CO2 + 2 mol H2O

From this, we can see that 1 mole of CH4 reacts with 2 moles of O2. Therefore, we can calculate the number of moles of O2 required as follows:

Number of moles of O2 = 1.19 mol CH4 x (2 mol O2 / 1 mol CH4) = 2.38 mol O2

Therefore, 19.04 grams of methane will react with 2.38 moles of oxygen gas.

Note: It's important to specify the substance and units in the answer, as requested.