Question:
$$\ce{2NH3 + CO2 -> (NH2)2CO + H2O}$$
If we have $637.2\ \mathrm{g}$ of $\ce{NH3}$ and $1142.0\ \mathrm{g}$ of $\ce{CO2}$ how many grams of $\ce{(NH2)2CO}$ will we get?
This is the explanation I was given:
First, convert reactives to moles:
- $637.2\ \mathrm{g}$ of $\ce{NH3}$ are $37.5$ moles.
- $1142.0\ \mathrm{g}$ of $\ce{CO2}$ are $26$ moles.
Second, define the stoichiometric proportion between reactives and products...
- From $2$ moles of $\ce{NH3}$ we get $1$ mol of $\ce{(NH2)2CO}$
- From $1$ mol of $\ce{CO2}$ we get $1$ mol of $\ce{(NH2)2CO}$
Third, calculate the number of product moles we would get if each reactive was completely consumed:
- Consuming all the $\ce{NH3}$ we would get $18.75$ moles of $\ce{(NH2)2CO}$
- Consuming all the $\ce{CO2}$ we would get $26$ moles of $\ce{(NH2)2CO}$
The limiting reagent is $\ce{NH3}$, so the most we will get of $\ce{(NH2)2CO}$ is $18.75$ moles. Now convert that to grams, we get $1125.0\ \mathrm{g}$.
- Why are we ignoring the existence of $\ce{H2O}$? Does it not matter?
- IF the formula is unbalanced, do I have to balance it first before making this calculation? What if I can't balance it?