First let's convert the 10.00 grams of H_{2}O into moles of H_{2}O. We do this by finding the molar mass of H_{2}O. Remember that the molar mass is the number of grams of a substance in 1 mole of the substance. So the molar mass of H_{2}O can be thought of as the number of grams of H_{2}O in 1 mole of H_{2}O.

So how do we find the molar mass of H_{2}O? We begin by looking on the periodic table to find the atomic masses of H and O. H is the symbol for hydrogen, and its atomic mass is 1.008. Remember that the H in H_{2}O has a subscript (the little number that is written right after the H). The subscript for H is 2. This means that we have 2 Hs in H_{2}O. The atomic mass for one H is 1.008 but we have 2 Hs in H_{2}O. So we have to do 2 x 1.008 in order to find the total grams of H in H_{2}O. The calculator says that 2 x 1.008 = 2.016. So we know that there are 2.016 grams of H in the molar mass of H_{2}O. Now we turn to the O in H_{2}O. O is the symbol for oxygen, and its atomic mass is 15.999. The O in H_{2}O has no subscript, so we know that there is only one O in H_{2}O. Now we add the grams of H and the grams of O together. We have 2.016 grams of H and 15.999 grams of O. If we add them together, the calculator says that 2.016 + 15.999 = 18.015 grams. So we know the molar mass of H_{2}O.

Molar mass of H_{2}O = 18.015 grams of H_{2}O in one mole of H_{2}O.

Remember that we found the molar mass of H_{2}O so that we could convert 10.00 grams of H_{2}O into moles of H_{2}O. We set up the equation to turn 10.00 grams of H_{2}O into moles of H_{2}O like this:

10.00 grams H_{2}O x __1 mole H___{2}__O__

18.015 grams H_{2}O

We solve this equation by doing 10.00/18.015. (That is 10 divided by 18.015 in case it is hard to read.) The calculator says that 10.00/18.015 = 0.55509. If you look in the equation above, you will see that we have grams on the top (in the "10.00 grams H_{2}O") and on the bottom (in the "18.015 grams H_{2}O"). This means that the grams cancel and vanish. So we are left with mole as the unit. And we already calculated

10/18.015 = 0.55509. When we add the unit of mole to this answer, we have:

0.55509 mole H_{2}O.

Now our amount of H_{2}O is ready to be put into an equation, so let's set it aside for a moment. Now let's look at the __coefficients__ in the equation we were given in the problem:

4 NH_{3} + 5 O_{2} --> 4NO + 6H_{2}O

The word coefficient refers to the number right before the compound or element. So in other words, NH_{3} has a coefficient of 4. O_{2} has a coefficient of 5. NO has a coefficient of 4, and H_{2}O has a coefficient of 6. These coefficients mean "the number of moles needed to make another substance". The coefficient of O_{2} is 5, and the coefficient of H_{2}O is 6, so we can interpret these coefficients as "we need 5 moles of O2 to produce 6 moles of H_{2}O". These coefficients are going to be very important in our next equation.

Let's look at how we set up our next equation. We start with the moles of H_{2}O and we want to get moles of O_{2}. Since we start with moles of H_{2}O, we put the coefficient of H_{2}O on the __bottom__ of the coefficients. This way, the word "H_{2}O" will cancel since it's on the top (in the "moles of H2O that we calculated") and on the bottom (in the "coefficient of H_{2}O"). We will be left with "moles of O_{2}" as our unit once we have put in the values and done the calculation. Our equation looks like this:

Moles of H_{2}O that we calculated x __coefficient of O___{2}

coefficient of H_{2}O

We know the moles of H_{2}O that we calculated. If you remember, we calculated that 10.00 grams of H_{2}O is equal to 0.55509 mole H_{2}O. So we put "0.55509 mole H_{2}O" in the equation instead of "moles of H_{2}O that we calculated". The equation now looks like this:

0.55509 mole H_{2}O x __coefficient of O___{2}

coefficient of H_{2}O

Now let's remember the coefficient of H_{2}O in the equation we were given. Remember that we found the coefficient of H_{2}O in the equation 4 NH_{3} + 5 O_{2} --> 4NO + 6H_{2}O. The coefficient for H_{2}O is 6. So we put 6 into our equation instead of "coefficient of H_{2}O". Our equation looks like this:

0.55509 mole H_{2}O x __coefficient of O___{2}

6

Now let's remember the coefficient of O_{2} in the equation we were given. Remember that we found the coefficient of O_{2} in the equation 4 NH_{3} + 5 O_{2} --> 4NO + 6H_{2}O. The coefficient for O_{2} is 5. So we put 5 into our equation instead of "coefficient of O_{2}". Our equation looks like this:

0.55509 mole H_{2}O x __5__

6

Now we can multiply 0.55509 x (5/6). The calculator says that 0.55509 x (5/6) = 0.46258. But what does this number mean? This is the moles of O_{2}. If the question had asked for the moles of O2, we could stop here. But the question asks us for the grams of O_{2}. So we have to convert 0.46258 moles of O2 to grams of O2.

How do we convert the moles of O_{2} to grams of O_{2}? We do this by finding the molar mass of O_{2}. To find the molar mass, we go to the periodic table and look for the atomic mass of O. The atomic mass of O is 15.999. Remember that this is the atomic mass of one O, but we have 2 Os in O_{2}. (We know because the O has a subscript of 2 in O_{2}). So we need to multiply 15.999 x 2 in order to find the molar mass of O_{2}. The calculator says that 15.999 x 2 = 31.998. Thus, we know:

Molar mass of O_{2} = 31.998 grams of O_{2} in one mole of O_{2}

Now we can find the number of grams of O_{2} in 0.46258 mole of O_{2}. We set up the equation like this:

0.46258 mole O_{2} x __31.998 grams O___{2}

1 mole O_{2}

We solve this equation by multiplying straight across. In other words, we do 0.46258 x 31.998. The calculator says that 0.46258 x 31.998 = 14.80. Notice that the "mole O_{2}" is on the top and the bottom, so it cancels out and we are left with the unit "grams O_{2}". So our answer is 14.80 grams of O_{2}.

**Final Answer:** 14.80 grams of O_{2} are required to make 10.00 grams of H_{2}O.

Kitty C.

Thank you so much for the detailed explanation, I know that must've taken a lot of time but I am very grateful and understand how to do this type of problem now!06/15/21