# Question 2a6da

##### 2 Answers
Mar 20, 2017

$\text{60.0 g}$

#### Explanation:

Notice that your reaction

${\text{N"_ (2(g)) + 3"H"_ (2(g)) -> color(blue)(2)"NH}}_{3 \left(g\right)}$

consumes $1$ mole of nitrogen and $3$ moles of hydrogen gas in order to produce $\textcolor{b l u e}{2}$ moles of ammonia.

You know that the reaction must produce $\text{73 g}$ of ammonia, so the first thing to do here is to convert the mass of the product to moles by using the compound's molar mass

73 color(red)(cancel(color(black)("g"))) * "1 mole NH"_3/(17.03color(red)(cancel(color(black)("g")))) = "4.287 moles NH"_3

You can now use the $1 : \textcolor{b l u e}{2}$ mole ratio that exists between nitrogen and ammonia to say that the reaction must consume

4.287 color(red)(cancel(color(black)("moles NH"_3))) * "1 mole N"_2/(color(blue)(2)color(red)(cancel(color(black)("moles NH"_3)))) = "2.1435 moles N"_2

in order to produce the needed amount of ammonia. To convert this to grams of nitrogen gas, use its molar mass

$2.1435 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles N"_2))) * "28.01 g"/(1color(red)(cancel(color(black)("mole N"_2)))) = color(darkgreen)(ul(color(black)("60.0 g}}}}$

As requested, the answer is rounded to one tenth of a gram, i.e. to one decimal place.

Mar 20, 2017

Calculate the number of moles required from the mass of ammonia and then calculate the mass of nitrogen in the correct molar ratio.

#### Explanation:

Use the balanced equation to relate the number of moles of ammonia to nitrogen. Calculate the number of moles required from the mass of ammonia and then calculate the mass of nitrogen in the correct molar ratio.

The balanced equation shows that one mole of nitrogen $\left({N}_{2}\right)$ will produce two moles of ammonia. 73 grams of ammonia is $\left(\frac{73}{17}\right) \cdot \left(\frac{g r a m s N {H}_{3}}{\frac{g r a m s}{m o l} N {H}_{3}}\right) = 4.29$ moles ammonia.

To produce this will require
((N_2)/(NH_3)) * (½) 4.29 = 2.15# moles nitrogen.

The equation uses the diatomic nitrogen form, so the correct “molecular weight” is 28g/mol.

$2.15 \text{moles} {N}_{2} \cdot 28 \left(\frac{g}{m o l}\right) = 60.1 g {N}_{2}$

NOTE that given only 73 as the initial value, correct number of significant figures in the answer should be only two also, or 60g.

The additional figure (60.1) requested is fine for intermediate work, and it is better to have an extra digit than too few. But be aware that it does NOT improve the inherent accuracy of the answer, which is still only 60g +/_ 0.5g at best.