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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.

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  • $\begingroup$ Thanks for the edits, I wasn't really sure how often to use LaTeX. Seems like whenever possible, good to know! $\endgroup$ – Tetramputechture Sep 9 '15 at 21:18
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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 $x$.

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  • $\begingroup$ Alright, that seems quite straightforward to me, which is why I was unsure when I started tackling the problem (I remember my professor taking more steps). Thank you! $\endgroup$ – Tetramputechture Sep 9 '15 at 21:24
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    $\begingroup$ @Tetramputechture You're welcome. Glad I helped $\endgroup$ – Oussama Boussif Sep 9 '15 at 21:27
  • $\begingroup$ Apparently the "correct" DE was dx/dt = k(50 - (x/5)(32 - (4x/5)). I just hope I don't get docked too many points for solving that wrong equation, too. :( $\endgroup$ – Tetramputechture Sep 15 '15 at 15:43

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