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A chemistry assignment has a student conduct a single replacement reaction by adding 7.5g of sodium metal to 20.0g of aluminum chloride. How much aluminum in moles would precipitate out as a result of the reaction

Sagot :

Snor1ax

Answer: 0.109 mol

Explanation:

Balanced chemical equation:

3 Na(s) + AlCl3(aq) --> 3 NaCl(aq) + Al(s)

3 moles of sodium metal react with 1 mole of aluminum chloride to produce 3 moles of aluminum chloride and 1 mole of solid aluminum

Molar mass of Na = 22.99 g/mol

Mass(Na) = 7.5 g

Find number of moles of Na:

n = mass of Na/molar mass of Na

=(7.5 g)/(22.99 g/mol)

= 0.3262 mol

Molar mass of AlCl3,

MM = 1*MM(Al) + 3*MM(Cl)

= 1*26.98 + 3*35.45

= 133.33 g/mol

Mass(AlCl3) = 20.0 g

Find number of moles of AlCl3,

n = mass of AlCl3/molar mass of AlCl3

=(20 g)/(1.333*10^2 g/mol)

= 0.15 mol

Balanced chemical equation is:

3 Na + AlCl3 ---> Al + 3 NaCl

3 mol of Na reacts with 1 mol of AlCl3

for 0.3262 mol of Na, 0.1087 mol of AlCl3 is required

But we have 0.15 mol of AlCl3

So, Na is the limiting reagent and we will use it in our further calculations.

According to balanced equation:

mol of Al formed = (1/3) * moles of Na

= (1/3)*0.3262

= 0.1087 mol

Answer: 0.109 mol of aluminum will precipitate out as a result of the reaction

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