Asked by lily
compute theoretical yield (g)
~2Al+3Cl2->2AlCl3
~2g Al, 2g Cl2
[ans:2.5g AlCl3]
i need the calculation work bcus i got diff from the ans which is 7.12 g.
~2Al+3Cl2->2AlCl3
~2g Al, 2g Cl2
[ans:2.5g AlCl3]
i need the calculation work bcus i got diff from the ans which is 7.12 g.
Answers
Answered by
DrBob222
moles Al = 2/26.98 = ??
moles Cl2 =2/70.9 = ??
moles Al x (2moles AlCl3/2 moles Al) = 0.0741 OR
moles Cl2 x (2 moles AlCl3/3 moles Cl2) = 0.0188
Therefore, the limiting reagent is Cl2, moles AlCl3 formed is 0.0188 and that x molar mass of approximately 133 = 2.5 g AlCl3 as theoretical yield.
moles Cl2 =2/70.9 = ??
moles Al x (2moles AlCl3/2 moles Al) = 0.0741 OR
moles Cl2 x (2 moles AlCl3/3 moles Cl2) = 0.0188
Therefore, the limiting reagent is Cl2, moles AlCl3 formed is 0.0188 and that x molar mass of approximately 133 = 2.5 g AlCl3 as theoretical yield.
Answered by
Anonymous
ME TOO. i was doing this question and got 7.4!! im so confused
Answered by
Yeeters Assemble
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