## anonymous one year ago Okay guys, I need help. Every time I've looked it up the answer is wrong...

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1. anonymous

For the reaction shown, compute the theoretical yield of product (in grams) for each of the following initial amounts of reactants. $2Al(s)+3Cl _{2}(g)→2AlCl _{3}(s)$

2. anonymous

1. 2.5 g Al, 2.5 g Cl2 2. 7.6 g Al, 24.6 g Cl2 3. 0.235 g Al, 1.20 g Cl2

3. anonymous

And the answer is NOT 6.2 or 6.3... Which is what I have been told like five times.

4. anonymous

Is 1, 2, and 3, the answer choices or what you're starting with?

5. anonymous

no, those are the "initial amounts of reactants shown" from the question. I just need one explained to me and I should get it.

6. anonymous

@roast_master_says @whpalmer4

7. anonymous

Ok, so if you have equal amounts of the reactants, you start like this: 2.5 g of Al $$\times$$ $$\sf \frac{1~mole~Al}{26.99~g~Al}$$ $$\times$$ $$\sf \frac{2~mol~AlCl_3}{2 ~mole~Al}$$ notice that the units should all cancel out! It's the same as math, when you have $$a$$ $$\times$$ $$\frac{C}{a}$$ = C beause the two $$a$$'s cancel out. Same here. treat them as variables. Do the same for Cl$$_2$$ 2.5 g of Al $$\times$$ $$\sf \frac{1~mole~Al}{26.99~g~Al}$$ $$\times$$ $$\sf \frac{2~mol~AlCl_3}{\color{red}{3 ~mole~Al}}$$ instead of 2, it's 3 moles.

8. anonymous

I understandthe ratio first step.. but every time I work it out I get 6.3.. Which is wrong.

9. aaronq

you have to identify the limiting reactant (LR) from each set of masses. convert the reactants to moles then divide by their respective stoichiometric coefficients from the balanced reaction. After this division, the species with the lower value is the LR. use the mass of the limiting reactant to carry out yield calculations. roastmaster has the right idea, but maybe this will make it clearer. Set up a ratio like this: $$\sf \dfrac{moles~of~LR}{LR's ~coefficient }=\dfrac{moles~of~product }{product's ~coefficient}$$ solve for moles of product with simple algebra $$\sf moles~of~product=\dfrac{moles~of~LR*product's ~coefficient}{LR's ~coefficient }$$ lastly convert the moles to mass using $$\sf moles=\dfrac{mass }{molar~mass}$$