When 85g of CH4 are mixed with 160g of O2 what is the maximum amount of CO2 that can be produced?
plis help me ASAP
2.50 moles
To find the maximum amount of CO2 that can be produced, we need to determine the limiting reactant in the given mixture of CH4 and O2.
First, let's calculate the moles of each substance. We'll use the molar mass of CH4 (methane) and O2 (oxygen) to do this.
The molar mass of CH4 is:
C: 12.01 g/mol x 1 = 12.01 g/mol
H: 1.008 g/mol x 4 = 4.032 g/mol
Total: 16.04 g/mol
The moles of CH4 can be calculated as follows:
85 g / 16.04 g/mol = 5.297 mol
Similarly, the molar mass of O2 is:
O: 16.00 g/mol x 2 = 32.00 g/mol
The moles of O2 can be calculated as follows:
160 g / 32.00 g/mol = 5.000 mol
Now, let's compare the mole ratio of CH4 to O2 using the balanced chemical equation for the combustion of methane:
CH4 + 2O2 -> CO2 + 2H2O
From the balanced equation, we see that 1 mole of CH4 reacts with 2 moles of O2 to produce 1 mole of CO2.
Since the mole ratio of CH4 to O2 is 1:2, and we have more moles of CH4 than O2 (5.297 mol vs. 5.000 mol), O2 is the limiting reactant.
Now, let's calculate the maximum amount of CO2 that can be produced using the moles of CH4.
Since 1 mole of CH4 reacts to produce 1 mole of CO2, the maximum amount of CO2 that can be produced is equal to the moles of CH4:
5.297 mol CH4 = 5.297 mol CO2
Finally, convert the amount of CO2 to grams using its molar mass. The molar mass of CO2 is:
C: 12.01 g/mol x 1 = 12.01 g/mol
O: 16.00 g/mol x 2 = 32.00 g/mol
The mass of CO2 is calculated as follows:
5.297 mol x (12.01 g/mol + 32.00 g/mol) = 224.30 g
Therefore, the maximum amount of CO2 that can be produced when 85g of CH4 is mixed with 160g of O2 is 224.30 grams.
This is a limiting reagent (LR) problem. You know that because amounts are given for BOTH reactants. Here is a step by step procedure for working LR problems. Print this procedure out.
1. Write and balance the equation.
CH4 + 2O2 ==> CO2+ 2H2O
2. Convert what you have to mols.
mols CH4 = grams/molar mass = ?
mols O2 = grams/molar mass = ?
3a. Using the coefficients in the balanced equation, convert mols CH4 to mols of either product.
b. Do the same and convert mols O2 to mols of the same product.
c. It is likely that these two values will not be the same which means one of them is not right. The correct value in LR problems is ALWAYS the smaller value.
4. Now convert the smaller number of mols of the product to grams. g = mols x molar mass