In a double replacement reaction between silver nitrate and magnesium bromide. Calculate the mass of silver bromide which should be produced from 22.5g of silver nitrate. The actual yield was 22.0g, how do you calculate percent yield?

Question

In a double replacement reaction between silver nitrate and magnesium bromide.  Calculate the mass of silver bromide which should be produced from 22.5g of silver nitrate.  The actual yield was 22.0g, how do you calculate percent yield?

Serenity Allen · Accepted Answer

Theoretical yield: #"24.9 g"#
Percent yield: #"88.4%"#

Start by writing the balanced chemical equation for this double replacement reaction

#2"AgNO"_text(3(aq]) + "MgBr"_text(2(aq]) -> color(red)(2)"AgBr"_text((s]) darr + "Mg"("NO"_3)_text(2(aq])#

Notice that you have a #1: color(red)(2)# mole ratio between silver nitrate and silver bromide. This means that the reaction will produce #color(red)(1)# mole of the latter for every #1# moel of the former that takes part in the reaction.

However, keep in mind that this is true for a reaction that has a #100%# yield - this represents the theoretical yield of the reaction.

If the percent yield of the reaction is smaller than #100%#, than silver bromide will not be produced in a #1:color(red)(1)# mole ratio with silver nitrate.

So, how many moles of silver nitrate do you have in #"22,5 g"# of silver nitrate? Use the compound's molar mass to get

#22.5color(red)(cancel(color(black)("g"))) * "1 mole AgNO"_3/(169.87color(red)(cancel(color(black)("g")))) = "0.13245 moles AgNO"_3#

So, what would the theoretical yield of the reaction be?

Well, the #1:1# mole ratio would produce

#0.13245color(red)(cancel(color(black)("moles AgNO"_3))) * "1 mole AgBr"/(1color(red)(cancel(color(black)("mole AgNO"_3)))) = "0.13245 moles AgBr"#

To calculate how many grams of silver bromide would contain this many moles? Once again, use the compound's molar mass

#0.13245color(red)(cancel(color(black)("moles"))) * "187.77 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("24.9 g AgBr")#

Now, a reaction's percent yield is defined as

#color(blue)(%"yield" = "actual yield"/"theoretical yield" xx 100)#

If the actual yield is #"22.0 g"#, then the percent yield of the reaction is

#"% yield" = (22.0color(red)(cancel(color(black)("g"))))/(24.9color(red)(cancel(color(black)("g")))) xx 100 = color(green)("88.4%")#

Here's a photo of how the silver bromide precipitate would look like

Cooper Anderson · Answer

undefined First, determine the balanced chemical equation for the reaction:

2AgNO₃ + MgBr₂ → 2AgBr + Mg(NO₃)₂

Next, calculate the molar mass of AgBr:
Ag = 107.87 g/mol
Br = 79.90 g/mol
Molar mass of AgBr = 107.87 g/mol + 79.90 g/mol = 187.77 g/mol

Now, calculate the theoretical yield of AgBr:
From the balanced equation, 2 moles of AgBr are produced for every 2 moles of AgNO₃ used.
So, the molar ratio of AgNO₃ to AgBr is 1:1.

Molar mass of AgNO₃ = 169.87 g/mol

Calculate moles of AgNO₃:
Moles = Mass / Molar mass
Moles = 22.5 g / 169.87 g/mol ≈ 0.1323 mol

Since the ratio of AgNO₃ to AgBr is 1:1, the moles of AgBr produced will also be 0.1323 mol.

Calculate the mass of AgBr:
Mass = Moles × Molar mass
Mass = 0.1323 mol × 187.77 g/mol ≈ 24.83 g

Now, calculate the percent yield:
Percent yield = (Actual yield / Theoretical yield) × 100
Percent yield = (22.0 g / 24.83 g) × 100 ≈ 88.6%

Therefore, the percent yield of silver bromide is approximately 88.6%.

Avery Adams · Answer

undefined To calculate the mass of silver bromide produced from 22.5g of silver nitrate, you first need to determine the stoichiometry of the reaction by balancing the chemical equation. Then, use stoichiometry to find the theoretical yield of silver bromide.

After finding the theoretical yield, you can calculate the percent yield using the formula:

$$
\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%
$$

Substitute the values for the actual yield (22.0g) and theoretical yield into the formula to find the percent yield.