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1.-A mixture of aluminum metal and chlorine gas reacts to form aluminum chloride (AlCl3): 2Al(s) +
3Cl2(g) → 2AlCl3(s). How many moles of aluminum chloride will form when 5 moles of chlorine gas react
with excess aluminum metal?

Respuesta :

Eduard22sly

The number of mole of aluminum chloride, AlCl₃ produced when 5 moles of chlorine gas, Cl₂ reacts is 3.33 moles

Balanced equation

2Al(s) + 3Cl₂(g) → 2AlCl₃(s)

From the balanced equation above,

3 moles of Cl₂ reacted to produce 2 moles of AlCl₃.

How to determine the moles of AlCl₃ produced

From the balanced equation above,

3 moles of Cl₂ reacted to produce 2 moles of AlCl₃.

Therefore,

5 moles of Cl₂ will react to = (5 × 2) / 3 = 3.33 moles of AlCl₃.

Thus, 3.33 moles of AlCl₃ were obtained from the reaction.

Learn more about stoichiometry:

https://brainly.com/question/14735801

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