[ThePeople]

You have a given reaction. You know the mass for one of the products and you shall estimate/calculate the mass for one of the original reactants. Give a general solution to the problem.

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I have no idea how to start, would anyone mind giving me a push into the right direction?

[Arkcon]

Hmmm....tricky.  OK, lets play along with what the question author possibly might want.  Try to write it out ... use x and y for the masses that you will know, pick other letters for the masses you won't.  Try to write a chemical equation, then simplify -- moving knowns to one side and unknowns to another.  Even so, this is a tough one.  Do they mean a single replacement reaction, a synthesis or decomposition reaction, a double replacement, or something more complicated?

[Borek]

You have a given reaction.

Given reaction means given stoichiometric coefficients, doesn't it?

[curiouscat]

You have a given reaction.

Given reaction means given stoichiometric coefficients, doesn't it?

I assumed so.

[curiouscat]

Hmmm....tricky.  OK, lets play along with what the question author possibly might want.  Try to write it out ... use x and y for the masses that you will know, pick other letters for the masses you won't.  Try to write a chemical equation, then simplify -- moving knowns to one side and unknowns to another.  Even so, this is a tough one.  Do they mean a single replacement reaction, a synthesis or decomposition reaction, a double replacement, or something more complicated?

Why isn't this simply

ν_r R + ...  ν_p P + ...

[tex]
m_r=\left( \frac{\nu_r}{\nu_p} \right ) \cdot \left ( \frac{MW_r}{MW_p} \right ) \cdot m_p \\
[/tex]

Is excess / limiting the issue?

[ThePeople]

Hmmm....tricky.  OK, lets play along with what the question author possibly might want.  Try to write it out ... use x and y for the masses that you will know, pick other letters for the masses you won't.  Try to write a chemical equation, then simplify -- moving knowns to one side and unknowns to another.  Even so, this is a tough one.  Do they mean a single replacement reaction, a synthesis or decomposition reaction, a double replacement, or something more complicated?

The lack of info disturbes me alot, I think my teacher wants to keep it simple. However, what we have studied for the past few days is biochemistry, chemical kinetics and stoichiometry. 

[curiouscat]

Tricky indeed. Here's another try to include Equilibrium effects:

Notation:
Suffix r for reactants (index j) and p for products (index i).
Choose any one reactant as denoted by suffix "ro"
ν denotes stoichiometric indices

First find ε based on given mass of one product
[tex]
\epsilon= \left ( \frac{m_P}{MW_P \cdot \nu_P}\right ) \\
[/tex]



Now solve this non-linear eqn. for the single variable n_r0
[tex]
K_{eq}=\frac{1}{(n_{r0}-\nu_{r0} \cdot \epsilon)^{\nu_{r0}}} \cdot \prod_{products,i}(\nu_{P_i} \cdot \epsilon )^{\nu_{P_i}} \cdot \frac{1}{\prod_{reactants,j,j \not= r0} \left( \frac{\nu_{rj}}{\nu_{ro}} \cdot n_{r0} - \nu_{rj} \cdot \epsilon \right)^{\nu_{rj}}}
[/tex]

Finally get the mass of reactant
[tex]
m_{r0}=n_{r0} \cdot MW_{ro}
[/tex]

This should work when all reactants are fed in stoichiometric proportions. I dont think there's a way to account for excesses.

Darn the formatting; not sure if the equations are legible. 

[ThePeople]

Tricky indeed. Here's another try to include Equilibrium effects:

Notation:
Suffix r for reactants (index j) and p for products (index i).
Choose any one reactant as denoted by suffix "ro"
ν denotes stoichiometric indices

First find ε based on given mass of one product
[tex]
\epsilon= \left ( \frac{m_P}{MW_P \cdot \nu_P}\right ) \\
[/tex]



Now solve this non-linear eqn. for the single variable n_r0
[tex]
K_{eq}=\frac{1}{(n_{r0}-\nu_{r0} \cdot \epsilon)^{\nu_{r0}}} \cdot \prod_{products,i}(\nu_{P_i} \cdot \epsilon )^{\nu_{P_i}} \cdot \frac{1}{\prod_{reactants,j,j \not= r0} \left( \frac{\nu_{rj}}{\nu_{ro}} \cdot n_{r0} - \nu_{rj} \cdot \epsilon \right)^{\nu_{rj}}}
[/tex]

Finally get the mass of reactant
[tex]
m_{r0}=n_{r0} \cdot MW_{ro}
[/tex]

This should work when all reactants are fed in stoichiometric proportions. I dont think there's a way to account for excesses.

Darn the formatting; not sure if the equations are legible. 

Holy f*ck....

[curiouscat]

Holy f*ck....

I'm not 100% sure it is right. So don't trust it. Maybe someone here will vett it / correct it.

Besides it looks more complicated than it really is because I tried to cram it into a single formula.