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Sonbull [250]
4 years ago
14

In the reaction of 1.23 g of salicylic acid, molar mass 138.12 g/mol, with 2.85 g of acetic anhydride, molar mass 102.10 g/mol,

a student obtained 1.39 g of acetylsalicylic acid, molar mass 180.17 g/mol. What is the percent yield?
I found that salicylic acid was the limiting reactant so do I just find the mass from there?
Chemistry
1 answer:
Rudik [331]4 years ago
7 0
The balanced reaction would be written as:
C7H6O3 + C4H6O3--->C9H8O4 + HC2H3O2
To determine the percent yield, we need to first determine the theoretical yield if the reaction were to proceed completely. Then, we divide the actual yield that is given to the theoretical yield times 100 percent. The limiting reactant from the reaction would be salicylic acid. We do as follows: 

Theoretical yield: 1.23 g C7H6O3 ( 1 mol / 138.21 g ) ( 1 mol C9H8O4 / 1 mol C7H6O3 ) ( 180.157 g / mol ) = 1.60 g <span>C9H8O4 should be produced

Percent yield = 1.39 / 1.60 x 100 = 86.88%

Thus, only 86.88% of the theoretical 
</span><span>C9H8O4 is being produced.</span>
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A 25.00 mL sample containing an unknown amount of Al3+ and Pb2+ required 17.14 mL of 0.04907 M EDTA to reach the end point. A 50
dolphi86 [110]

Answer:

pAl³⁺ = 1,699

pPb²⁺ = 1,866

Explanation:

In this problem, the first titration with EDTA gives the moles of Al³⁺ and Pb²⁺, these moles are:

Al³⁺ + Pb²⁺ in 25,00mL = 0,04907M×0,01714L = <em>8,411x10⁻⁴ moles of EDTA≡ moles of Al³⁺ + Pb²⁺.</em>

The molar concentration is <em>8,411x10⁻⁴ moles/0,02500L = </em><em>0,0336M Al³⁺+Pb²⁺.</em>

In the second part each Al³⁺ reacts with F⁻ to form AlF₃. Thus, you will have in solution just Pb²⁺.

The moles added of EDTA are:

0,02500L×0,04907M = 1,227x10⁻³ moles of EDTA

The moles of EDTA in excess that react with Mn²⁺ are:

0,02064M × 0,0265L = 5,470x10⁻⁴ moles of Mn²⁺≡ moles of EDTA

That means that moles of EDTA that reacted with Pb²⁺ are:

1,227x10⁻³ moles - 5,470x10⁻⁴ moles = 6,800x10⁻⁴ moles of EDTA ≡ moles of Pb²⁺.

The molar concentration of Pb²⁺ is:

6,800x10⁻⁴mol/0,0500L = <em>0,0136 M Pb²⁺</em>

Thus, molar concentration of Al³⁺ is:

0,0336M Al³⁺+Pb²⁺ - 0,0136 M Pb²⁺ = <em>0,0200M Al³⁺</em>

pM is -log[M], thus pAl³⁺ and pPb²⁺ are:

<em>pAl³⁺ = 1,699</em>

<em>pPb²⁺ = 1,866</em>

I hope it helps!

3 0
3 years ago
(4Ga + 3S2 → 2Ga2S3)
nexus9112 [7]

Answer:

n_S=1.076molS

Explanation:

Hello,

In this case, given the undergoing chemical reaction, we can see a 4:3 mole ratio between the consumed moles of gallium and sulfur respectively, therefore, the consumed moles of sulfur, from the 100.0 g of gallium (use its atomic mass) turn out:

n_{S}=100.0gGa*\frac{1molGa}{69.72gGa}*\frac{3molS}{4molS}  \\\\n_S=1.076molS

Best regards.

7 0
3 years ago
Why is it not possible to always have a controlled experiment
lesantik [10]
A controlled experiment is one in which evrerything is held constant except for one verieble, maybe is usually a st of data is taken for a control group.
4 0
3 years ago
Is it posible Count a grain of sand?explain
Alla [95]

Answer:

Yes

Explanation:

As long as its a solid, you can count it. It will be hard, but possible.

5 0
3 years ago
Six moles of an ideal gas are in a cylinder fitted at one end with a movable piston. the initial temperature of the gas is 27.0c
Alla [95]

We have to know final temperature of the gas after it has done 2.40 X 10³ Joule of work.

The final temperature is: 75.11 °C.

The work done at constant pressure, W=nR(T₂-T₁)

n= number of moles of gases=6 (Given), R=Molar gas constant, T₂= Final temperature in Kelvin, T₁= Initial temperature in Kelvin =27°C or 300 K (Given).

W=2.4 × 10³ Joule (Given)

From the expression,

(T₂-T₁)=\frac{W}{nR}

(T₂-T₁)= \frac{2.40 X 10^{3} }{6 X 8.314}

(T₂-T₁)= 48.11

T₂=300+48.11=348.11 K= 75.11 °C

Final temperature is 75.11 °C.


6 0
3 years ago
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