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Yakvenalex [24]
3 years ago
8

A chemist determined by measurements that 0.050 moles of aluminum participated in a chemical reaction. Calculate the mass of alu

minum that participated in the chemical reaction. Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
Law Incorporation [45]3 years ago
8 0

Answer:

13.5 g

Explanation:

This question is solved easily if we remember that the number of moles is obtained by dividing the mass into the atomic weight or molar mass depending if we are referring to elements or molecules.

Therefore, the mass of aluminum in the reaction will the 0.050 mol Al times the atomic weight of aluminum.

number of moles = n = mass of Al / Atomic Weight Al

⇒ mass Al = n x Atomic Weight Al = 0.050 mol x 27 g mol⁻¹

                                                         = 13.5 g

We have three significant figures in 0.050 and therefore we should have three significant figures in our answer.

           

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faltersainse [42]

1.39 g HCl

Explanation:

The balanced chemical equation for this reaction is given by

Zn(<em>s</em>) + 2HCl(<em>aq</em>) ---> ZnCl2(<em>aq</em>) + H2(<em>g</em>)

Convert the # of grams of Zn to moles:

1.25 g Zn × (1 mol Zn/65.38 g Zn) = 0.0191 mol Zn

Use the molar ratio to find the # of moles of HCl needed to react completely with the given amount of Zn:

0.0191 mol Zn × (2 mol HCl/1 mol Zn) = 0.0382 mol HCl

Convert this amount to grams:

0.0382 mol HCl × (36.458 g HCl/1 mol HCl) = 1.39 g HCl

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2 years ago
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Este último electrón estará en un orbital de 2pz como lo muestran los números cuánticos enumerados anteriormente.

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