# Question 1efb5

Mar 16, 2015

$3 {H}_{2 \left(g\right)} + {N}_{2 \left(g\right)} \to 2 N {H}_{3 \left(g\right)}$

This first thing you must do is have a good look at the mole ratios that exist between all the species involved in the reaction. Notice that you need 3 moles of hydrogen gas to react with 1 mole of nitrogen gas in order to produce 2 moles of ammonia.

These ratios are the key to this problem. So, you know that, when 12.3 g of hydrogen react with excess nitrogen, 55.8 g of ammonia are formed.

You need to check to see what the percent yield of this reaction is, i.e. how much of the reactants are turned into the product.

The number of moles of hydrogen that reacted is

${\text{12.3 g H"_2 * "1 mole"/"2.02 g" = "6.089 moles H}}_{2}$

According to the $\text{3:2}$ mole ratio that exists between hydrogen and ammonia, that much hydrogen should have produced

${\text{6.089 moles H"_2 * "2 moles NH"_3/"3 moles H"_2 = "4.0593 moles NH}}_{3}$

However, the reaction actually produced

${\text{55.8 g NH"_3 * "1 mole NH"_3/"17.03 g" = "3.2766 moles NH}}_{3}$

This means that, as suspected, your reaction does not have a 100% yield; the actual yield is

$\text{% yield" = "actual yield"/"theoretical yield} \cdot 100$

$\text{% yield" = "3.2766 moles"/"4.0593 moles" * 100 = "80.7%}$

The percent yield of the reaction is very important because it will tell you exactly how much ammonia will be produced when 280.2 g of nitrogen react with excess hydrogen.

This time, the number of moles of nitrogen that react will be

${\text{280.2 g N"_2 * "1 mole N"_2/"28.01 g" = "10.00 moles N}}_{2}$

According to the $\text{1:2}$ mole ratio that exists between nitrogen and ammonia, this much nitrogen should have produced

${\text{10.00 moles N"_2 * "2 moles NH"_3/"1 mole N"_2 = "20.00 moles NH}}_{3}$

But since the reaction's percent yield is not 100%, but actually 80.7%, that much nitrogen will produce

$\text{% yield" = "actual yield"/"theoretical yield} \cdot 100$

"actual yield" = ("% yield" * "theoretical yield")/100#

${\text{actual yield" = (80.7 * 20.00)/100 = "16.14 moles NH}}_{3}$

Use ammonia's molar mass to determine the mass produced

${\text{16.14 moles NH"_3 * "17.03 g"/"1 mole NH"_3 = "274.86 g NH}}_{3}$

Since you know the temperature (I won't write the conversion to Kelvin since the answer is quite long already), volume, and number of moles, you can use the ideal gas law equation to solve for the system's pressure

$P V = n R T \implies P = \frac{n R T}{V}$

$P = \left(\text{16.14 moles" * 0.082("atm" * "L")/("mol" * "K") * ("273.15 K"))/("2.00 L}\right)$

$P = \text{194.11 atm}$

Rounded to three sig figs, the answer will be

$P = \text{194 atm}$