Two chemicals $\mathrm{X}$ and $\mathrm{Y}$ react in such a way that $1\,\rm g$ of $\mathrm{X}$ combines with $4\,\rm g$ of $\mathrm{Y}$ to produce a compound $\mathrm{Z}$ which is written as:
$$\mathrm{X + 4Y \to Z}$$
The rate at which $\mathrm{X}$ and $\mathrm{Y}$ react (the rate at which $\mathrm{Z}$ is formed) is proportional to the product of the amount of $\mathrm{X}$ and $\mathrm{Y}$ that remained untransformed at that time. Assume there is initally $50\,\rm g$ of $\mathrm{X}$ and $32\,\rm g$ of $\mathrm{Y}$, and it is known that $30\,\rm g$ of $\mathrm{Z}$ are formed in $10\,\rm min$.
Derive a differential equation initial value problem (sic) for $Z(t)$.
Right now, I have this, where $\alpha (t)$ represents the transformed $\mathrm{X}$ at time $t$, and $\beta (t)$ represents the transformed $\mathrm{Y}$ at time $t$:
$$\frac{\Bbb dZ}{\Bbb dt} = k\big(50 - \alpha (t)\big) \big(32 - \beta (t)\big)$$
Where do I go from here? I know I'll have to use the given relation, and eventually have to only use a $Z$ instead of $\alpha$ and $\beta$, but how?
Once I get a proper separable equation, I should be good to go.
Let $x$ be the quantity of $\mathrm{Z}$ that was formed. So, the remaining quantities of each component are (I assume you know this as it pure chemistry):
$$ \begin{align} n_\mathrm{X}&=50-x\\ n_\mathrm{Y}&=32-4x\\ n_\mathrm{Z}&=x \end{align} $$
And, since the rate at which $\mathrm{Z}$ reacts is proportional to the quantities you mentioned, then:
$$ \frac{dn_\mathrm{Z}}{dt}=\frac{dx}{dt}=k(50-x)(32-4x) $$
I assume that's easy and straightforward. You can now separate it and solve for $x$.