# Question 292bf

Dec 21, 2017

The limiting reactant is ${\text{NH}}_{3}$.

#### Explanation:

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.

Step 1. Gather all the information in one place with molar masses above the formulas and everything else below the formulas.

${M}_{r} : \textcolor{w h i t e}{m m m m l l} 17.03 \textcolor{w h i t e}{m l l} 44.01$
$\textcolor{w h i t e}{m m m m m m m} \text{2NH"_3 +"CO"_2 color(white)(l)→ ("NH")_2"CO" + "H"_2"O}$
$\text{Mass/g} : \textcolor{w h i t e}{m m l l} 637.2 \textcolor{w h i t e}{m l l} 1142$
$\text{Amt/mol:} \textcolor{w h i t e}{m m} 37.42 \textcolor{w h i t e}{m l l} 25.95$
$\text{Divide by:} \textcolor{w h i t e}{m m m} 2 \textcolor{w h i t e}{m m m l l} 1$
$\text{Moles rxn:} \textcolor{w h i t e}{m l l} 18.71 \textcolor{w h i t e}{m l l} 25.95$

${\text{Moles of NH"_3 = 637.2color(red)(cancel(color(black)("g NH"_3))) × ("1 mol NH"_3)/(17.03 color(red)(cancel(color(black)("g NH"_3)))) = "37.42 mol NH}}_{3}$

$\text{Moles of CO"_2 = 1142 color(red)(cancel(color(black)("g NH"_3))) × ("1 mol Fe")/(55.84 color(red)(cancel(color(black)("g Fe")))) = "25.95 mol Fe}$

Step 2. Identify the limiting reactant

An easy way to identify the limiting reactant is to calculate the "moles of reaction" each will give:

You divide the moles of each reactant by its corresponding coefficient in the balanced equation.

I did that for you in the table above.

${\text{NH}}_{3}$ is the limiting reactant because it gives fewer moles of reaction.

Dec 21, 2017

The limiting reactant is $N {H}_{3}$

#### Explanation:

1. Find the molar masses of the involved compounds which are obtainable from the periodic table.
$N {H}_{3} = \frac{17 g}{m o l}$
$C {O}_{2} = \frac{44 g}{m o l}$
2. Given the masses of two reactants, find each value in moles. Convert mass to mole through molar conversion which conversion factors are obtainable from its respective molar masses shown above; i.e.,
$N {H}_{3}$:
$= 637.2 \cancel{g N {H}_{3}} \times \frac{1 m o l N {H}_{3}}{17 \cancel{g N {H}_{3}}} = 37.4824 m o l$
$C {O}_{2}$:
$= 1142 \cancel{g C {O}_{2}} \times \frac{1 m o l C {O}_{2}}{44 \cancel{g C {O}_{2}}} = 25.9545 m o l$
3. Analyze data based on the amount of moles needed by each reactant for complete reaction. Assuming: the production of urea ${\left(N {H}_{2}\right)}_{2} C O$. Refer to the given balanced equation for the mole ratios.
color(red)(NH_3:
$= 37.4824 \cancel{m o l N {H}_{3}} \times \frac{1 m o l C {O}_{2}}{2 \cancel{m o l N {H}_{3}}}$
$= 18.7412 m o l C {O}_{2}$
This means that $\textcolor{red}{37.4824 m o l N {H}_{3} \text{ needs } 18.7412 m o l C {O}_{2}}$ for complete reaction.
$\text{Therefore} :$
$\frac{C {O}_{2} \text{ available ")/(25.9545molCO_2)>(CO_2 " required }}{18.7412 m o l C {O}_{2}}$
color(red)(NH_3 " is the limiting reactant"
color(blue)(CO_2:
$= 25.9545 \cancel{m o l C {O}_{2}} \times \frac{2 m o l N {H}_{3}}{1 \cancel{m o l C {O}_{2}}}$
$= 51.9090 m o l N {H}_{3}$
This means that color(blue)(25.9545molCO_2 " needs " 51.9090molNH_3 for complete reaction.
$\text{Therefore} :$
$\frac{N {H}_{3} \text{ available ")/(37.4824molNH_3)<(NH_3 " required }}{51.9090 m o l N {H}_{3}}$
color(blue)(CO_2 " is the excess reactant"#
4. Therefore, the limiting reactant is $N {H}_{3}$ in the production of urea (carbamide).