Can you determine the mass of carbon monxide that is produced when 45.6 grams of methane reacts with 73.2 grams of oxygen gas. Can you please show work in detail please.
Thanks
This is a limiting reagent problem (actually that is two problems in one). All problems of this type are done the same way. Print this out and memorize how to do it.
1. Write and balance the equation.
2CH4 + 3O2 ==> 2CO + 4H2O
2a. Convert grams CH4 to moles. moles = grams/molar mass.
2b. Do the same for oxygen.
3a. Using the coefficients in the balanced equation, convert moles CH4 to moles of the product.
3b. Do the same for moles oxygen.
3c. It is quite likely that the answer form 3a and 3b will be different which means one of them is wrong. The correct answer, in limiting reagent problems, is ALWAYS the smaller value.
4. Convert the moles from 3c to grams of the product. grams = moles x molar mass
To determine the mass of carbon monoxide produced when methane reacts with oxygen gas, we need to first write a balanced chemical equation for the reaction. The balanced equation for the combustion of methane is:
CH₄ + 2O₂ → CO₂ + 2H₂O
From the equation, we can see that 1 mole of methane (CH₄) reacts with 2 moles of oxygen gas (O₂) to produce 1 mole of carbon monoxide (CO) and 2 moles of water (H₂O).
To find the moles of methane and oxygen gas, we can use the molar masses of the substances. The molar mass of methane (CH₄) is 16 g/mol, and the molar mass of oxygen gas (O₂) is 32 g/mol.
Moles of methane = mass of methane / molar mass of methane
Moles of methane = 45.6 g / 16 g/mol
Moles of methane = 2.85 mol
Moles of oxygen gas = mass of oxygen gas / molar mass of oxygen gas
Moles of oxygen gas = 73.2 g / 32 g/mol
Moles of oxygen gas = 2.29 mol
According to the balanced equation, the reaction requires 1 mole of methane to react with 2 moles of oxygen gas. However, we have excess oxygen gas, meaning we have more oxygen gas than needed. The limiting reactant in this case is methane since there is less of it compared to oxygen gas. Thus, the moles of CO produced will be according to the moles of methane used.
From the balanced equation, 1 mole of methane produces 1 mole of carbon monoxide. Therefore, we can conclude that the moles of carbon monoxide produced will also be 2.85 mol.
To find the mass of carbon monoxide produced, we can use the molar mass of carbon monoxide (CO), which is 28 g/mol.
Mass of carbon monoxide = moles of carbon monoxide × molar mass of carbon monoxide
Mass of carbon monoxide = 2.85 mol × 28 g/mol
Mass of carbon monoxide = 79.8 g
Hence, the mass of carbon monoxide produced when 45.6 grams of methane reacts with 73.2 grams of oxygen gas is 79.8 grams.