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2 years ago

Please help with an easy percentage yield question ASAP!!! Would really appreciate!

Chemistry

1 answer:

Brut [27]2 years ago

7 0

Answer:

The mass percentage yield is 85%

Explanation:

Step 1: The balanced equation

2 C7H6O3 + (CH3CO)2CO → 2 C9H8O4 + H20

Step 2: Given data

mass of salicylic acid = 1.03 grams

Volume of acetic anhydride = 2.00 mL = 2*10^-3 L

The product is 1.16 g aspirin synthesized

Molar mass of salicylic acid = 138.12 g/mole

Molar mass of aspirin = 180.16 g/mole

Density of acetic anhydride = 1.0820 g/mL

Step 3: Calculated moles of salicylic acid

Number of moles of salicylic acid = mass of salicylic acid / Molar mass of salicylic acid

Number of moles of salicylic acid = 1.03 grams / 138.12 g/mole

Number of moles of salicylic acid =0.0075 moles

Step 4: Calculated mass acetic anhydride

mass of acetic anhydride = 1.0820 g/mL * 2 mL = 2.164 grams

Step 5: Calculate number of moles of acetic anhydride

Number of moles = 2.164 grams / 102.09 g/moles

Number of moles = 0.0212 moles

Step 6: Find amount of reacting moles

C7H6O3 is the limiting reactant, there will react 0.0075 moles

Since for 2 moles C7H6O3 consumed, we need 1 mole of (CH3CO)2CO to produce 2 moles of C9H804

So for 0.0075 moles C7H6O3 consumed, we produce 0.0075 moles of C9H8O4

Step 7: Calculate mass of aspirin

mass of aspirin = 0.0075 moles * 180 g/mole = 1.35 grams

Step 8: Calculate the mass percentage yield

(1.16 grams / 1.35 grams) * 100% = 85 %

The mass percentage yield is 85%

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If 15.6 grams of copper (ii) chloride react with 20.2 grams of sodium nitrate how many grams of sodium chloride can be formed? W

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Answer:

- 13.56 g of sodium chloride are theoretically yielded.

- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.

- 0.50 g of sodium nitrate remain when the reaction stops.

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Explanation:

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In this case, according to the question, it is possible to set up the following chemical reaction:

$CuCl_2+2NaNO_3\rightarrow 2NaCl+Cu(NO_3)_2$

Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:

$m_{NaCl}^{by\ CuCl_2}=15.6gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2} *\frac{2molNaCl}{1molCuCl_2} *\frac{58.44gNaCl}{1molNaCl} =13.56gNaCl\\\\m_{NaCl}^{by\ NaNO_3}=20.2gNaNO_3*\frac{1molNaNO_3}{84.99gNaNO_3} *\frac{2molNaCl}{2molNaNO_3} *\frac{58.44gNaCl}{1molNaCl} =13.89gNaCl$

Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:

$m_{NaNO_3}^{by\ NaCl}=13.56gNaCl*\frac{1molNaCl}{58.44gNaCl}*\frac{2molNaNO_3}{2molNaCl} *\frac{84.99gNaNO_3}{1molNaNO_3}=19.72gNaNO_3$

Therefore, the leftover of sodium nitrate is:

$m_{NaNO_3}^{leftover}=20.2g-19.7g=0.5gNaNO_3$

Finally, the percent yield is computed via:

$Y=\frac{12.6g}{13.56g} *100\%\\\\Y=92.9\%$

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2 years ago

Read 2 more answers

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butalik [34]

Answer:

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Explanation:

The first part of the question asks to convert the mass of the calcium carbonate given to number of moles.

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Number of moles = mass/molar mass

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So the number of moles of CaCO3 will be 2.49/100 = 0.0249 moles

The second part of the question asks to convert the mass of carbon iv oxide to moles of carbon iv oxide

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That is same as ;

Number of moles = mass/molar mass

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Number of moles of CO2 = 1.13/44 = 0.0256 moles

Now, if we compare the values of these number of moles, we can see that there are almost equal.

What this means is that the number of moles of calcium carbonate reacted is equal to the number of moles of carbon iv oxide produced.

So what we conclude here is that we have an equal mole ratio between the two compounds.

So the reaction that would be the correct answer will present equal number of moles of carbon iv oxide and calcium carbonate

Thus, we can see that reaction B is the one that governs this process as it is the only reaction out of the three options that present the two compounds with equal number of moles.

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