Question

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

Answer 1

0.15 moles of aluminum would precipitate out as a result of the reaction between sodium metal and aluminum chloride.

Step-by-step explanation:

To determine how much aluminum in moles would precipitate out as a result of the reaction between sodium metal and aluminum chloride, we first need to balance the chemical equation:

2 Na + 2 AlCl3 → 3 NaCl + 2 Al

From the balanced equation, we can see that 2 moles of aluminum chloride react with 2 moles of sodium to produce 2 moles of aluminum. This means that for every 2 moles of aluminum chloride, 2 moles of aluminum would precipitate out.

Given that the student added 7.5g of sodium metal and 20.0g of aluminum chloride, we can calculate the number of moles of aluminum chloride:

Number of moles of aluminum chloride = mass of aluminum chloride / molar mass of aluminum chloride

Using the molar masses of sodium (22.99 g/mol) and aluminum chloride (133.34 g/mol), we can convert the mass of aluminum chloride to moles and then determine the amount of aluminum that would precipitate out.

Substituting the values into the formula:

Number of moles of aluminum chloride = 20.0g / 133.34 g/mol = 0.15 moles of aluminum chloride

Since the stoichiometric ratio from the balanced equation indicates that 2 moles of aluminum chloride react with 2 moles of sodium to produce 2 moles of aluminum, we can conclude that 0.15 moles of aluminum chloride would result in 0.15 moles of aluminum precipitating out.

Answer 2

Answer: 0.109 mol

Step-by-step 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