When 15.0 g of methane (CH₄) burns completely in oxygen, what is the theoretical yield of water (H₂O)? Then, if the actual yield is 28.5 g, find the percent yield.

Divide by coefficients: Al: ( 0.9266 / 2 = 0.4633 ) Cl₂: ( 0.4231 / 3 = 0.1410 ) → Cl₂ is limiting.

2Al (s) + 3Cl2 (g) → 2AlCl3 (s)

If 5.0 g of Al and 7.0 g of Cl2 are reacted?

What is the limiting reagent in the reaction:

( \frac35.037.62 \times 100% = 93.0% )

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