Step by step chemistry

Question

A student mixed 24.5 g of magnesium bromide with 8.67 x 1022 molecules of hydroiodic acid... They recovered 8.23 g of acid. What is the percent yield?

Answer 1

70.7% yield

Let me know if this explanation below helps!

Problem Roadmap:

1) Find moles of each substance

2) Identify limiting reagent

3) Identify produced acid

4) Identify theoretical yield and actual yield in grams.

5) Determine % yield

The balanced chemical equation is:

MgBr₂ + 2HI → MgI₂ + 2HBr

Let's start with finding the moles of MgBr₂ and HI given.

HI: Avogadro's number, the amount of molecules in 1 mole, is 6.022*10²³. The amount of moles of hydroiodic acid (HI) from 8.67*10²² molecules is:

(8.67*10²² molecules HI)/(6.022*10²³ molecules/mole) = 0.143972 moles HI

MgBr₂: Its molar mass is 184.113 g/mol. We are given 24.5 g of this molecule. So, the amount of moles of magnesium bromide (MgBr₂) would be solved by dividing the given mass by the known molar mass:

24.5g MgBr₂ / 184.113 g/mol MgBr₂ = 0.13307 moles MgBr₂

Now let's use the mole ratio to find the limiting reagent.

Remember the equation: MgBr₂ + 2HI → MgI₂ + 2HBr. There is a mole ratio between MgBr₂ and HI, shown by the coefficients, of 1 mol MgBr₂ : 2 mol HI .

We have 0.13307 moles MgBr₂ and 0.143972 moles HI. This is not nearly enough HI for the amount of MgBr₂ given, so HI is the limiting reagent. Now, we should try finding a ratio between this limiting reagent and the produced acid.

Determining the produced acid identity and the HI : acid ratio.

Once again, we look back to the equation MgBr₂ + 2HI → MgI₂ + 2HBr, but this time we are looking at the product side to find the new acid. Since HBr is the only molecule of the 2 that has a proton to donate, HBr is the produced acid. Looking at the 2 coefficient on HI and the 2 coefficient on the HBr, there is a HI : Br ratio of 1 mol HI : 1 mol HBr.

Theoretical and Actual Yields

Theoretical:

We know there is a 1:1 molar ratio between HI and HBr, and we also found that there is 0.143972 moles of HI being reacted. So, theoretically, 0.143972 moles of HBr should be produced. HBr has a molar mass of 80.91g. So, by multiplying the theoretical moles by the molar mass, we get:

0.143972 mol HBr*80.91 g/mol HBr = 11.6488 g HBr produced theoretically

Actual:

We are given that 8.23g of the acid HBr is produced and recovered from the reaction, so we are already given the actual yield.

Finally, finding the % yield

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

Substituting our numbers, we have: (8.23g/11.6488g) * 100% = 70.65% --> 70.7% yield (respecting the 3 given significant figures)