Hello, Mar,
Thanks for providing answers! I find the second answer is correct, but the first is WAY out of the park. I'm assuming you wanted help on how those answers were obtained? If so, read on.
Q1. What is the volume of H2O produced from a reaction of 5.98g of O2? (Answer = 8.372 L of H2O)
2H2 + O2 → 2H2O
[Update: Please check the Comments section for a revision to my answer to this question. If one assumes a gas phase volume at STP, then the answer of 8.372 L of H2O is correct. ]
One's first reaction to the answer is "That's a LOT of water from only 5.98 grams of O2, I wonder if the unit should have been ml instead of L (or kg instead of g). There are only 0.187 moles of O2, which should result in twice that amount of H2O, according to the balanced equation. The 0.374 moles of water represent only 6.73 grams of water. Assuming a density of water of 1g/ml, this is 6.73 ml of water, not 8.372 liters. 6.73 grams makes sense to me. Conservation of matter means we should get a total water mass of slightly more than the amount of O2 we put in. Since hydrogen is such a puny small atom, it doesn't add much to the final mass (apologies, H2).
[See comments for a correction]
Q2. How many moles of NaCl are produced from 12.45L of Cl2? (Answer = 1.11 moles of Cl2)
2Na + Cl2 → 2NaCl
The answer of 1.11 moles Cl2 is correct, if one assumes STP for the Cl2 gas. We need STP to use the 22.4 liters/mole conversion factor for all gases at STP. The balanced equation says we need twice as many moles of Na than Cl2, so the last step is easy, once we have moles Cl2 (0.556 moles) we multiply by 2. 1.11 moles Na matches your suggested answer.
Bob
Robert S.
01/10/21
J.R. S.
01/10/21