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Phantasy [73]
2 years ago
9

Solve 21 and 22 please

Chemistry
1 answer:
Nutka1998 [239]2 years ago
7 0

Answer:

try youre best to succsess

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A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s
Alla [95]

<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For NaBr:</u>

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol

The chemical equation for the reaction of 1-butanol and NaBr is:

\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = \frac{1}{1}\times 0.108=0.108 moles of 1-bromobutane

  • Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g

  • To calculate the percentage yield of 1-bromobutane, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%

Hence, the percent yield of the 1-bromobutane is 48.65 %

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What is the difference between creep and a landslide?
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4 0
3 years ago
1.86 g H2 is allowed to react with 9.75 g N2 , producing 2.87g NH3.
svet-max [94.6K]

Answer:

                     (a)  Theoretical Yield  =  10.50 g

                      (b)   %age yield  = 27.33 %

Explanation:

Answer-Part-(a)

                 The balance chemical equation for the synthesis of Ammonia is as follow;

                                          N₂ + 3 H₂ → 2 NH₃

Step 1: Calculating moles of N₂ as;

                   Moles = Mass / M/Mass

                   Moles = 9.75 g / 28.01 g/mol

                   Moles = 0.348 moles of N₂

Step 2: Calculating moles of H₂ as;

                   Moles = Mass / M/Mass

                   Moles = 1.86 g / 2.01 g/mol

                   Moles = 0.925 moles

Step 3: Finding Limiting reagent as;

According to equation,

                1 mole of N₂ reacts with  =  3 moles of H₂

So,

             0.348 moles of N₂ will react with  =  X moles of H₂

Solving for X,

                     X = 3 mol × 0.348 mol / 1 mol

                     X = 1.044 mol of H₂

It shows that to consume 0.348 moles of N₂ completely we require 1.044 mol of Hydrogen while, as given in statement we are only provided with 0.925 moles of H₂ hence, hydrogen  is limiting reagent. Therefore, H₂ will control the final yield.

Step 4: Calculating moles of Ammonia as,

According to equation,

                3 mole of H₂ produces  =  2 moles of NH₃

So,

             0.925 moles of H₂ will produce  =  X moles of NH₃

Solving for X,

                     X = 2 mol × 0.925 mol / 3 mol

                     X = 0.616 mol of NH₃

Step 5: Calculating theoretical yield of Ammonia as,

                     Theoretical Yield  =  Moles × M.Mass

                     Theoretical Yield  =  0.616 mol  × 17.03 g/mol

                     Theoretical Yield  =  10.50 g

Answer-Part-(b)

                    %age yield  = Actual Yield / Theoretical Yield × 100

                    %age yield  = 2.87 g / 10.50 g × 100

                    %age yield  = 27.33 %

4 0
3 years ago
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