Question

A 0.156 g piece of solid aluminum reacts with gaseous oxygen from the atmosphere to form solid aluminum oxide. In the laboratory, a student weighs the mass of the aluminum oxide collected from this reaction as 0.189 g.

What is the percent yield of this reaction?

Answer 1

The percent yield of this reaction is 64.1 %

Step-by-step explanation:

Step 1: Data given

Mass of solid aluminium = 0.156 grams

Mass of aluminium oxide produced = 0.189 grams

Atomic mass of aluminium = 26.98 g/mol

Molar mass of aluminium oxide = 101.96 g/mol

Step 2: The balanced equation

4Al + 3O2 → 2Al2O3

Step 3: Calculate moles aluminium

Moles aluminium = mass aluminium / molar mass aluminium

Moles aluminium = 0.156 grams / 26.98 g/mol

Moles aluminium = 0.00578 moles

Step 4: Calculate moles aluminium oxide

For 4 moles aluminium we need 3 moles O2 to produce 2 moles Al2O3

For 0.00578 moles aluminium we'll have 0.00578/2 =0.00289 moles aluminium oxide

Step 5: Calculate mass of aluminium oxide

Mass Al2O3 = moles Al2O3 * molar mass Al2O3

Mass Al2O3 = 0.00289 moles * 101.96 grams

Mass Al2O3 = 0.295 grams

Step 6: Calculate percent yield

% yield = (actual yield / percent yield) * 100%

% yield = (0.189 grams / 0.295 grams) * 100 %

% yield = 64.1 %

The percent yield of this reaction is 64.1 %

Answer 2

The percent yield of this reaction is 64.3 %

Step-by-step explanation:

First of all we determine the reation:

4Al(s) + 3O₂(g) → 2Al₂O₃

2nd step: we determine the moles ( mass / molar mass )

0.156 g / 26.98 g/mol = 0.00578 moles

We assume O₂ as the excess reagent so the limiting is the Al

Ratio is 4:2 so now we make a rule of three

4 Al produce 2 moles of Al₂O₃

Then 0.00578 moles of Al must produce ( 0.00578 . 2) /4 = 0.00289 moles of oxide.

These moles are the 100 % yield reaction. Let' s convert the moles to mass

0.00289 mol . 101.96 g / mol = 0.294 g

This is the way to calculate the percent yield:

(Produced yield / Theoretical Yield ) . 100 : (0.189 g /0.294 g) .100 = 64.3 %