Answer:
To determine the number of grams of H2O produced when 20 grams of CH4 reacts with excess O2, we need to use stoichiometry.
First, we need to find the molar mass of CH4 (methane). The molar mass of carbon (C) is approximately 12.01 grams per mole, and the molar mass of hydrogen (H) is approximately 1.01 grams per mole. Since there is only one carbon atom and four hydrogen atoms in CH4, the molar mass of CH4 is 12.01 + (4 × 1.01) = 16.05 grams per mole.
Next, we convert the given mass of CH4 to moles by dividing the mass by the molar mass. So, 20 grams of CH4 equals 20 / 16.05 ≈ 1.244 moles.
Let's use the balanced equation to find the molar ratio between CH4 and H2O. According to the equation, 1 mole of CH4 reacts to produce 2 moles of H2O.
Therefore, using the molar ratio, we can calculate the number of moles of H2O produced. Since we have 1.244 moles of CH4, we can multiply it by the molar ratio of H2O to CH4: 1.244 moles CH4 × (2 moles H2O / 1 mole CH4) = 2.488 moles H2O.
Finally, we need to convert the moles of H2O to grams by multiplying the number of moles by the molar mass of H2O. The molar group of oxygen (O) is approximately 16.00 grams per mole, and the molar mass of hydrogen (H) is approximately 1.01 grams per mole. Since there are two hydrogen atoms and one oxygen atom in H2O, the molar group of H2O is (2 × 1.01) + 16.00 = 18.02 grams per mole.
Thus, 2.488 moles of H2O × 18.02 grams per mole ≈ of 44.87 grams of H2O should be produced.
To summarize:
When 20 grams of CH4 reacts with excess O2, approximately 44.87 grams of H2O should be produced.
Step-by-step explanation: