Question

5g of hydrogen react with 33 gram of oxygen .. what is mass of water formed? which is the limiting reagent?

Answer 1

We need to calculate this in moles. So 1 mole is equal to the atomic weight of a substance in grams, and each mole will have \(6.022\times10^{23}\) atoms. 1gram of hydrogen equals 1 mole, and 18 grams of oxygen equals one mole. So available we have 5 moles of hydrogen, and 33/16 = 2.0625 moles of Oxygen.

To make water we need two hydrogen atom for every oxygen atom. So immediately we can see that the limiting reagent will be the oxygen, but lets show why.

We have 5 moles of hydrogen, which makes up a total of \(5\times 6.022\times10^{23}=3.011\times10^{24}\) atoms. We have 2.0625 moles of Oxygen which makes a total of \(2.0625\times 6.022\times10^{23}=1.242\times10^{24}\) atoms. So clearly there are less atoms of oxygen than hydrogen. Whats more since we need 2 hydrogen per oxygen, then we need \(2.484\times10^{24}\) hydrogen, leaving about a third of the hydrogen remaining with nothing to react with.

But what is the mass of the water. We have 2.0625 moles of oxygen, and hence we need 4.125 moles of hydrogen. We know that 2.0625 moles of Oxygen has a mass of 33 grams. We also know that 1 mole of hydrogen ways 1 gram. hence 4.125 moles of hydrogen weigh 4.125 grams. Hence the final mass of the water is 4.125+33= 37.125 grams.

Answer 2

I ended up with the same answer thereabouts but with a different method. Worth noting that 1 mol of Oxygen is approx 16g, not 18g.
First a balanced equation:
\[2H _{2}+O _{2}\rightarrow2H _{2}O\]
Find out the number of moles provided:
\[5.00gH _{2}(1mol H _{2}/2.0158g H _{2})=2.48 mol H _{2}\]
\[33.0gO(1mol O/30.9988g O)=1.03 mol O\]
\[1.03 mol O_{2} (2 mol H_{2}/1 mol O_{2})=2.06 mol H_{2} required\]
O2 is the limiting reagent. Now we can calculate the moles of water:
\[1.03 mol O_{2}(2 mol H_{2}O/1mol O_{2})=2.06 mol H_{2}O\]
\[2.06 mol H_{2}O(18.0152g H_{2}O/1 mol H_{2}O)=37.158g H_{2}O\]
Which in correct SF is 37.2g H2O which is about the same answer provided above.

Answer 3

Slight typo in Osirisis post. 33.0gO(1mol O/30.9988g O)=1.03 mol O should read 33.0gO(1mol O/31.9988g O)=1.03 mol O.

And difference in the above answer from mine comes from Osirisis using the more accurate values for molar mass, as opposed to my approximate values.

My statement of 18 grams per mole of oxygen, is a typo, you will see that I use the value of 16 grams for the calculation.

Answer 4

Yes, you're right. Missed the 2 on the Oxygen on the second equation. The reason I used the O2 and H2 is they're found paired in nature. That's how my chem class had us approach the problems. I figured you had a typo. Your answer was close to mine from a SF perspective.