According to the law of conservation of mass, how many grams of water will be produced in the above reaction? If you are unsure how to work out this problem, see a similar example linked HERE. a 54 grams b 61 grams c 8 grams d 68 grams
c 8 grams
The law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction. In this reaction, 36 grams of methane (CH4) react with 96 grams of oxygen (O2) to produce carbon dioxide (CO2) and water (H2O).
To calculate the mass of water produced, we need to find the molar masses of each compound involved in the reaction:
- CH4: 12.01 g/mol (C) + 4(1.01 g/mol) = 16.05 g/mol
- O2: 2(16.00 g/mol) = 32.00 g/mol
- CO2: 12.01 g/mol (C) + 2(16.00 g/mol) = 44.01 g/mol
- H2O: 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
Next, we need to determine the limiting reactant in the reaction. We do this by calculating the moles of each reactant:
- Moles of CH4 = 36 g / 16.05 g/mol = 2.24 mol
- Moles of O2 = 96 g / 32.00 g/mol = 3.00 mol
The balanced chemical equation shows that 2 moles of water are produced for every 1 mole of methane reacted. Since there are 2.24 moles of methane, the maximum amount of water that can be produced is 2.24 x 2 = 4.48 moles.
Finally, we can calculate the mass of water produced:
- Mass of water = 4.48 mol x 18.02 g/mol = 80.54 grams
Therefore, the correct answer is c) 8 grams.