A chemist burns 160.0 g of Al in O to produce Al2O3. She produces 260.0 g of Al2O3. Calculate the theoritical yield and percent yield

Answers

Answer 1
Answer:

Answer:

86.04 %

Explanation:

Data given:

mass of Al  = 160.0 g

actual yield Al₂O₃= 260 g

Theoretical yield = ?

percent yield ofAl₂O₃ = ?

Solution:

First we have to find theoretical yield.

So,

we Look at the balanced reaction

             4Al  + 3O₂ -----—> 2Al₂O₃

             4 mol                      2 mol

As 4 mole of Al give 2 mole of Al₂O₃

Convert moles to mass

molar mass of Al = 27 g/mol

molar mass of Al₂O₃ = 2(27) + 3(16)

molar mass of Al₂O₃ = 54 + 48

molar mass of Al₂O₃ = 102 g/mol

Now

             4Al         +       3O₂   -----—>     2Al₂O₃

        4 mol (27g/mol)                       2 mol (102 g/mol)

             108 g                                            204 g

108 grams of Al produce 204 g of Al₂O₃

So

if 108 grams of Al produce 204 g of Al₂O₃ so how many grams of Al₂O₃will be produced by 160 g of Al.

Apply Unity Formula

              108 grams of Al  ≅ 204 g of Al₂O₃

              160 grams of Al ≅ X of Al₂O₃

Do cross multiply

              mass of Al₂O₃= 204 g x 160 g / 108 g

              mass of Al₂O₃ = 302.2 g

So the Theoretical yield of Al₂O₃ = 302.2 g

Now Find the percent yield of Al₂O₃

Formula Used

             percent yield = actual yield /theoretical yield x 100 %

Put value in the above formula

            percent yield = 260g / 302.2 g x 100 %

            percent yield = 86.04 %

percent yield of Al₂O₃ = 86.04 %


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47.0 ml of an HBr solution were rotated with 37.5 ml of a 0.215 m lioh solution to reach the equivalent point. What is the molarity of the hbr solution

Answers

Answer:

0.172 mol·L⁻¹

Step-by-step explanation:

1. Write the chemical equation for the reaction.

HBr + LiOH ⟶ LiBr + H₂O

===============

2. Calculate the moles of LiOH

c = n/V         Multiply each side by V

n = Vc

V = 0.0375 L

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Moles of LiOH = 0.0375× 0.215

Moles of LiOH = 8.062 × 10⁻³ mol LiOH

===============

3. Calculate the moles of HBr

1 mol HBr ≡ 1 mol LiOH

Moles of HBr = 8.062 × 10⁻³× 1/1

Moles of HBr = 8.062 × 10⁻³ mol HBr

===============

4. Calculate the molar concentration of the HBr

V = 47.0 mL = 0.0470 L

c = 8.062 × 10⁻³/0.0470

c = 0.172 mol·L⁻¹

The molarity of the HBr solution is 0,172 M.

The neutralization reaction between LiOH and HBr is the following:

HBr(aq) + LiOH(aq) → LiBr(aq) + H₂O(l)