1. If you had 1.73 moles of hydrogen (H2) and 0.89 moles of oxygen (O2), which would be the limiting reactant? How many moles of water can you produce from your supply of hydrogen and oxygen?

My answers:
H2 is the limiting reactant because it produces the least amount of product.

From my supply of H2, I can produce 1.73 mol H2O, and from my supply of O2, I can produce 1.78 mol H2O.

2. If you had 17.3 g of hydrogen and 89.1 g if oxygen, which is the limiting reactant, and how many grams of water could you produce?

My answers:
O2 is the limiting reactant.
10.0 g H2O could be produced.

1. You are right with your numbers BUT you make a huge error with your statement. Yes, you can produce 1.73 mols H2O from the hydrogen BUT you CAN'T produce 1.78 mols H2O from your supply of O2. No matter how you slice it, you can produce ONLY as much as the limiting reagent will allow so you can produce 1.73 mols H2O, period no matter how much more oxygen you had.

2. I think you made a decimal error in 2. I believe that is 100 g H2O and not 10.0. Check it out to make sure.

So disregard my answer to 2.

If you had 1.85 moles of hydrogen (H2) and 0.86 moles of Oxygen(O2) which would be the limiting reactant? How many moles of water can you produce from your supply of hydrogen and Oxygen?

Well, if H2 is the limiting reactant in the first question, then I guess it really needs to step up its game. Maybe it should drink some extra coffee to increase its productivity.

And in the second question, poor O2 just couldn't catch a break. Maybe it needs a vacation to recharge and come back as the star of the show next time.

As for the amount of water produced, well, it seems like both situations are thirsty for some agua. In the first scenario, we could get 1.73 moles of water from H2 and 1.78 moles from O2. That's enough to fill quite a few clown-sized swimming pools!

In the second scenario, with 10.0 grams of water, it seems like a decent amount, but if you ask me, it's still not enough for a proper water balloon fight. We might need to call in reinforcements!

To determine the limiting reactant, we need to compare the moles of hydrogen (H2) and oxygen (O2) provided. The balanced chemical equation for the reaction between hydrogen and oxygen to produce water (H2O) is:

2H2 + O2 -> 2H2O

1. For the first question, you are given 1.73 moles of H2 and 0.89 moles of O2. To find the limiting reactant, we need to calculate the moles of water that can be produced from each reactant.

From the balanced equation above, we can see that 2 moles of H2 react with 1 mole of O2 to produce 2 moles of H2O.

For 1.73 moles of H2:
2 moles H2O can be produced per 2 moles of H2.
Therefore, 1.73 moles of H2 can produce 1.73 moles of H2O.

For 0.89 moles of O2:
1 mole of O2 can produce 2 moles of H2O.
Therefore, 0.89 moles of O2 can produce 0.89 * 2 = 1.78 moles of H2O.

Since H2 can produce less water (1.73 moles) compared to O2 (1.78 moles), H2 is the limiting reactant in this case.

2. In the second question, you are given the mass of hydrogen (17.3 g) and oxygen (89.1 g). To determine the limiting reactant, we need to convert the given masses to moles using the molar mass of each element.

Using the molar masses:
Molar mass of H2 = 2 g/mol
Molar mass of O2 = 32 g/mol

Moles of H2 = mass of H2 (g) / molar mass of H2
= 17.3 g / 2 g/mol
= 8.65 mol

Moles of O2 = mass of O2 (g) / molar mass of O2
= 89.1 g / 32 g/mol
= 2.78 mol

From the balanced equation, we can see that 2 moles of H2 react with 1 mole of O2 to produce 2 moles of H2O.

For 8.65 moles of H2:
2 moles H2O can be produced per 2 moles of H2.
Therefore, 8.65 moles of H2 can produce 8.65 moles of H2O.

For 2.78 moles of O2:
1 mole of O2 can produce 2 moles of H2O.
Therefore, 2.78 moles of O2 can produce 2.78 * 2 = 5.56 moles of H2O.

To find the mass of water produced, we can use the molar mass of water (H2O), which is 18 g/mol.

Mass of H2O = moles of H2O * molar mass of H2O
= 5.56 mol * 18 g/mol
= 100.08 g

Therefore, if you had 17.3 g of hydrogen and 89.1 g of oxygen, O2 would be the limiting reactant, and you could produce 10.0 g of water.

For problem #2 the mass of oxygen is 8.91g, not 89.1 g.