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the reaction of aluminum with chlorine gas is shown 2Al + 3Cl2 -> 2AlCl3 based on this equation how many molecules of chlorin
1 answer:
45 molecules of chlorine gas (Cl₂) are needed to react with 30 atoms of aluminum (Al)
The balanced equation for the reaction is given below:
2Al + 3Cl₂ —> 2AlCl₃
From the balanced equation above,
2 atoms of Al required 3 molecules of Cl₂.
With the above information, we can determine the number of molecules of Cl₂ needed to react with 30 atoms of Al. This can be obtained as follow:
From the balanced equation above,
2 atoms of Al required 3 molecules of Cl₂.
Therefore,
30 atoms of Al will require = = 45 molecules of Cl₂.
Thus, 45 molecules of chlorine gas (Cl₂) are needed to react with 30 atoms of aluminum (Al)
Learn more: brainly.com/question/24918379
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Answer:
0.591 g of magnesium phosphate is the theoretical yield.
Magnesium nitrate is the limiting reactant.
Explanation:
Hello!
In this case, since the balanced reaction turns out:
Next, we compute the grams of magnesium phosphate yielded by each reactant, considering the present mole ratios and molar masses:
Thus, we infer that the correct theoretical yielded mass is 0.591 g as magnesium nitrate is the limiting reactant for which it produces the fewest grams of product.
However, is not possible to compute the percent yield since no actual yield is given, and must be provided or indicated by the problem or an experiment and it not here, nevertheless, you may compute the percent yield by dividing the actual yield by the theoretical and then multiplying by 100:
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