When 85.0 g of CH4 are mixed with 160. g of O2 the limiting reactant is __________.

CH4 + 2O2 �¨ CO2 + 2H2O
A) CH4
B) O2
C) CO2
D) H2O

O2

O2

CH4 + O2 ==> CO2 + 2H2O

mols CH4 = grams/molar mass = ?
mols O2 = grams/molar mass = ?

Convert mols CH4 to mols CO2 using the coefficients in the balanced equation.
Do the same with O2.
It is likely that these two values for mols CO2 will not agree which means one of them is wrong. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.

Well, let's do some math here. To find the limiting reactant, we need to compare the amount of moles in each reactant.

The molar mass of CH4 is approximately 16 g/mol, so 85.0 g of CH4 is approximately 85.0 g / 16 g/mol = 5.31 moles of CH4.

The molar mass of O2 is approximately 32 g/mol, so 160. g of O2 is approximately 160. g / 32 g/mol = 5.00 moles of O2.

Since we have more moles of CH4 (5.31 moles) than O2 (5.00 moles), the limiting reactant is O2. So, the answer is B) O2.

Why do you think O2 is the limiting reactant? Because it's tired of always being told to "go take a deep breath" and it has decided to take a break. Can you blame it?

To find the limiting reactant in this chemical reaction, we need to compare the number of moles of each reactant to determine which one will be completely consumed first.

First, we need to convert the masses of CH4 and O2 to moles. We can do this by dividing the mass of each substance by its molar mass.

The molar mass of CH4 (methane) is calculated as:
1 carbon (C) + 4 hydrogen (H) = 12.01 g/mol + (4 * 1.01 g/mol) = 16.04 g/mol

So, the number of moles of CH4 can be calculated by dividing the mass of CH4 by its molar mass:
Number of moles of CH4 = 85.0 g / 16.04 g/mol = 5.3 mol

Next, we need to calculate the number of moles of O2. The molar mass of O2 (oxygen) is 32.00 g/mol.

Number of moles of O2 = 160.0 g / 32.00 g/mol = 5.0 mol

Now, we can compare the ratios of the coefficients of CH4 and O2 in the balanced equation to determine which reactant is limiting.

The balanced equation shows that the ratio of CH4 to O2 is 1:2.

From the calculations, we have:
Number of moles of CH4 = 5.3 mol
Number of moles of O2 = 5.0 mol

Since the ratio of CH4 to O2 is 1:2, it means that we need twice as much O2 as CH4.

Therefore, since we have more CH4 (5.3 mol) than the stoichiometrically required amount (5.0 mol), the limiting reactant in this reaction is O2.

Hence, the correct answer is B) O2.