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?

The percent yield of this reaction is 64.1 %

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 %

The percent yield of this reaction is 64.3 %

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 %