# Question 2a580

Jun 20, 2017

Here's how you can do that.

#### Explanation:

First of all, reactants are not produced by the reaction, they react, hence the name, to produce products.

Now, the thing to remember about converting between moles and grams is that our tool of choice is the molar mass of the compound.

The molar mass represents the mass of exactly $1$ mole of a given substance.

$\text{molar mass" = "mass in grams"/"1 mole}$

This means that in order to convert from moles to grams, you just need to multiply the number of moles by the molar mass

color(red)(cancel(color(black)("moles"))) * "mass in grams"/(1color(red)(cancel(color(black)("mole")))) = "mass in grams"

Similarly, to convert from grams to moles, simply flip the conversion factor and multiply

color(red)(cancel(color(black)("grams"))) * "1 mole"/("mass"color(white)(.) color(red)(cancel(color(black)("in grams")))) = "moles"

Now, let's say that you want to convert $3$ moles of ammonia, ${\text{NH}}_{3}$, to grams. Ammonia has a molar mass of ${\text{17.031 g mol}}^{- 1}$, which emans that $1$ mole of ammonia has a mass of $\text{17.031 g}$.

You will thus have

${\text{molar mass NH"_3 = "17.031 g"/"1 mole NH}}_{3}$

This means that the mass of $3$ moles of ammonia is equal to

3 color(red)(cancel(color(black)("moles NH"_3))) * "17.031 g"/(1color(red)(cancel(color(black)("mole NH"_3)))) = "51.093 g"

Similarly, the mass of $0.75$ moles of water, knowing that water has a molar mass of ${\text{18.015 g mol}}^{- 1}$, will be equal to

0.75 color(red)(cancel(color(black)("moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "13.51 g"#

So remember, the molar mass can be used as conversion factor to go from

• $\text{moles " -> " grams: " "mass in grams"/"1 mole}$
• $\text{grams " -> " moles: " "1 mole"/"mass in grams}$

Jun 20, 2017

So, for example, lets say we react 100 g of $N {H}_{3}$. Molar mass of $N {H}_{3}$ is 17 g/mol, so 100 g is (100/17) = 5.88 moles. The equation says that for every 4 moles of $N {H}_{3}$ that reacts you produce 4 moles of $N O$. Therefore you will produce 5.88 moles of $N O$. Molar mass of $N O$ is 44 g/mol, so you will produce 5.88 x 44 = 258.72 g of $N O$.