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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For the reaction shown, compute the theoretical yield of product (in grams) for each of the following initial amounts of reactants. \[2Al(s)+3Cl _{2}(g)→2AlCl _{3}(s)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.01. 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And the answer is NOT 6.2 or 6.3... Which is what I have been told like five times.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is 1, 2, and 3, the answer choices or what you're starting with?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no, those are the "initial amounts of reactants shown" from the question. I just need one explained to me and I should get it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@roast_master_says @whpalmer4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok, 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I understandthe ratio first step.. but every time I work it out I get 6.3.. Which is wrong.

aaronq
 one year ago
Best ResponseYou've already chosen the best response.1you 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 \(\sf moles=\dfrac{mass }{molar~mass}\)
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