16g of CH4 reacts with 64g of O2 producing 44g of CO2 How many grams of water are produced?
To solve this, we need to first identify the balanced chemical equation for the reaction between methane (CH4) and oxygen (O2) to produce carbon dioxide (CO2) and water (H2O).
The balanced chemical equation is:
CH4 + 2O2 → CO2 + 2H2O
From the equation, we see that for every 1 mole of CH4, we need 2 moles of O2 to produce 1 mole of CO2 and 2 moles of H2O.
Given that 16g of CH4 reacts with 64g of O2 and produces 44g of CO2, we can calculate the amount of water produced as follows:
1. Calculate the number of moles of CH4:
Molar mass of CH4 = 12g/mol (C) + 4g/mol (H) = 16g/mol (CH4)
Number of moles of CH4 = 16g / 16g/mol = 1 mole
2. Calculate the number of moles of O2 using the mole ratio from the balanced equation:
1 mole of CH4 reacts with 2 moles of O2
Number of moles of O2 = 64g / 32g/mol = 2 moles
3. Determine the limiting reactant (CH4 in this case), by comparing the moles of CH4 and O2. Since CH4 is the limiting reactant, it will be completely consumed.
4. Using the mole ratio from the balanced equation, determine the number of moles of H2O produced:
1 mole of CH4 produces 2 moles of H2O
Number of moles of H2O = 1 mole CH4 x 2 moles H2O/mole CH4 = 2 moles
5. Calculate the mass of water produced:
Molar mass of H2O = 2g/mol (H) + 16g/mol (O) = 18g/mol (H2O)
Mass of water produced = 2 moles x 18g/mol = 36g
Therefore, 36g of water are produced in the reaction.