The reaction of methane and oxygen yields carbon dioxide gas and water. The unbalanced reaction is as follows: CH4(g) + O2(g)CO2(g) + H2O(g)
A. If 1.26 grams of methane are reacted, how many grams of water vapor are produced?
B. If 2.796 grams CO2 are released, how many grams of methane were reacted?
1. Balance the equation.
2. Convert 1.26 g CH4 to moles. mols = grams/molar mass.
3. Using the coefficients in the balanced equation, convert moles CH4 to moles H2O.
4. Convert moles H2O to grams H2O. grams = mols x molar mass.
B is done the same way.
Post your work if you get stuck.
I don,t know about question.!!!!
To solve these problems, we can use stoichiometry, which is based on the balanced equation of the reaction. Let's first balance the equation:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
A. To find the number of grams of water vapor produced, we need to use the molar ratio between methane and water. The molar ratio is 1:2, which means for every 1 mole of methane reacted, 2 moles of water vapor are produced.
1 mole of CH4 = 16.04 g (molar mass of CH4)
2 moles of H2O = 2 × 18.02 g = 36.04 g (molar mass of H2O)
Now, let's calculate the amount of water vapor produced:
1.26 g CH4 * (2 moles H2O / 1 mole CH4) * (36.04 g / 2 moles H2O) = 22.87 g
Therefore, 1.26 grams of methane will produce approximately 22.87 grams of water vapor.
B. To find the number of grams of methane reacted, we need to use the molar ratio between methane and CO2. The molar ratio is 1:1, which means for every 1 mole of methane reacted, 1 mole of CO2 is produced.
1 mole of CH4 = 16.04 g (molar mass of CH4)
1 mole of CO2 = 44.01 g (molar mass of CO2)
Now, let's calculate the amount of methane reacted:
2.796 g CO2 * (1 mole CH4 / 1 mole CO2) * (16.04 g / 1 mole CH4) = 44.33 g
Therefore, 2.796 grams of CO2 were produced by reacting approximately 44.33 grams of methane.
A. To find the grams of water vapor produced, you need to use stoichiometry. Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction.
First, we need to balance the chemical equation:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Now, we can set up a proportion using the balanced equation:
1 mol CH4 → 2 mol H2O
To find the number of moles of CH4, we can use the molar mass of CH4:
1 mol CH4 = 16.04 g CH4
So, 1.26 g CH4 is equal to:
1.26 g CH4 × (1 mol CH4 / 16.04 g CH4) = 0.0786 mol CH4
According to the balanced equation, 1 mol of CH4 produces 2 mol of H2O. Therefore, multiplying the moles of CH4 by the ratio of H2O to CH4, we get the moles of H2O:
0.0786 mol CH4 × (2 mol H2O / 1 mol CH4) = 0.1572 mol H2O
Finally, to convert the moles of H2O to grams, we use the molar mass of water (H2O):
1 mol H2O = 18.02 g H2O
So, the number of grams of water vapor produced is:
0.1572 mol H2O × (18.02 g H2O / 1 mol H2O) ≈ 2.83 g H2O
Therefore, approximately 2.83 grams of water vapor are produced when 1.26 grams of methane are reacted.
B. To find the grams of methane reacted, we use stoichiometry again.
Using the balanced equation: CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
We know that 1 mol of CH4 reacts to produce 1 mol of CO2. So, the number of moles of CO2 produced is equal to the number of moles of CH4 reacted.
Given that the molar mass of CO2 is 44.01 g/mol, we can convert the grams of CO2 to moles:
2.796 g CO2 × (1 mol CO2 / 44.01 g CO2) ≈ 0.0635 mol CO2
Since 1 mol of CH4 produces 1 mol of CO2, the number of moles of CH4 is also 0.0635 mol.
Using the molar mass of CH4 (16.04 g/mol), we can convert the moles of CH4 to grams:
0.0635 mol CH4 × (16.04 g CH4 / 1 mol CH4) ≈ 1.02 g CH4
Therefore, approximately 1.02 grams of methane were reacted to produce 2.796 grams of CO2.