Percentage Yield

Question

A chemist reacts 100 g of PbCl₄ with excess NH₄Cl. They obtain an 87% yield (NH₄)₂PbCl₆. How many grams did they obtain?

Answer 1

A chemist reacts 100 g of PbCl₄ with excess NH₄Cl. They obtain an 87% yield (NH₄)₂PbCl₆. How many grams did they obtain?

Write the balanced equation PbCl₄ + 2NH₄Cl ====> (NH₄)₂PbCl₆.

The calculation used to determine the full (100%) yield of (NH₄)₂PbCl₆. is

Mol.. Mass (NH₄)₂PbCl₆. / Mol. Mass PbCl₄ x 100 g PbCl₄ = 456.0 g / 349.0 g x 100 g = 130.7 g of product.

But the %yield was 87%, so the mass of NH₄)₂PbCl₆.obtained is 130.7 x 0.87 = 114 g.

The molecular masses can be calculated using the periodic table.

Please check the math.

Check math.

Answer 2

Hi Po!

In order to answer this question, we first need a balanced equation for the reaction given to us:

PbCl₄ + NH₄Cl —> (NH₄)₂PbCl₆

left side: right side:

Pb: 1 Pb: 1 balanced

Cl: 4 +1 = 5 Cl: 6 unbalanced

N: 1 N: 2 unbalanced

H: 4 H: 8 unbalanced

To balance this equation, we can add a coefficient of 2 to the reactant NH₄Cl, giving us:

PbCl₄ + 2NH₄Cl —> (NH₄)₂PbCl₆

left side: right side:

Pb: 1 Pb: 1 balanced

Cl: 4 +2 = 6 Cl: 6 balanced

N: 2 N: 2 balanced

H: 8 H: 8 balanced

Now we are ready to use dimensional analysis to calculate the expected yield:

PbCl₄ + 2NH₄Cl —> (NH₄)₂PbCl₆

↓ ↑

100g → → → → (?)g

Important note: we are starting with the mass of PbCl₄, since it was reacted with excess NH₄Cl. This means that PbCl₄ is our limiting reactant.

[100g PbCl₄] [1 mol PbCl₄] [1 mol (NH₄)₂PbCl₆] [453.962g (NH₄)₂PbCl₆]

---------------- * ------------------- * ---------------------------- * --------------------------------

1 [349g PbCl₄] [1 mol PbCl₄] [1 mol (NH₄)₂PbCl₆]

After we cancel out units and multiply across the top and bottom of the fractions, we have:

[100 * 453.962]

------------------------ = 130.07 g (NH₄)₂PbCl₆ **this is our expected yield*

349

To calculate our actual yield, which we were told was 87% of the expected yield, we multiply our answer from the previous step by 87%, or 0.87:

130.07 * (0.87) = 113.165 g (NH₄)₂PbCl₆ **Remember to round using sig figs if needed here!**

Hope this helps!