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Stells [14]
2 years ago
15

An experiment reacts 20.4 g of zinc metal with a solution containing an excess of iron (III) sulfate. After the reaction, 10.8 g

rams of iron metal are recovered. What is the percent yield of the experiment?
Chemistry
2 answers:
quester [9]2 years ago
5 0

Answer:

The answer to your question is 92.7%

Explanation:

Balanced Chemical reaction

                             3 Zn  + Fe₂(SO₄)₃   ⇒   2Fe   +   3ZnSO₄

Molecular weight

Zinc = 65.4 x 3 = 196.2g

Iron (III) = 56 x 2 = 112 g

Proportions  

                           196.2 g of Zinc ------------------ 112 g of Iron

                            20.4 g of Zinc  -----------------   x

                            x = (20.4 x 112) / 196.2

                            x = 2284.8/196.2

                            x = 11.65 g of Iron

% yield = \frac{10.8}{11.65}  x 100

% yield = 0.927 x 100

% yield = 92.7

Anvisha [2.4K]2 years ago
4 0
Answer:

93.1%

Explanation:

As with all stoichometry problems, you must start by writing a balanced equation.

3 Zn (s) + Fe2O3 (s) → 2 Fe (s) + 3 ZnO (s)

Since you are given that there is 20.4g of Zinc metal as a reactant, that is your given. You will use this number to calculate your theoretical yield by using the mole ratio between Zinc and Iron.

20.4 g Zn • 1 mol Zn/65.38 g Zn • 2 mol Fe/3 mol Zn • 55.85g Fe/1 mol Fe = 11.6g Fe

Therefore, your theoretical yield is 11.6g.

Now, you divide your actual yield (10.8g) by your theoretical yield (11.6g) and multiply by 100 to get a percentage.

10.8g/11.6g • 100 = 93.1%

Therefore, your percent yield is 93.1%.
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How many moles are in 2.8x10^23 atoms of Calcium?
Zielflug [23.3K]

Answer: 1 mole ➡️ 6.022×10²³ atoms of si.

X mole ➡️ 2.8×10²⁴ atoms of si.

X = 2.8×10×10²³/6.022×10²³

= 28/6.022

= 4.65 moles.

Explanation:

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3 years ago
A fifth-grade science class puts half of a white tablet in a test tube with 10 milliliters (mL) of water at 21°C (70°F). The tab
hichkok12 [17]

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change in water tempreture.

Explanation:

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2 years ago
How many grams of oxygen are required too burn 8.8g of c3h8?
hichkok12 [17]
From the balanced equation: 
<span>1mol C3H8 requires 5mol O2 for combustion </span>
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<span>8.8g C3H8 = 8.8/44 = 0.2mol C3H8 </span>
<span>This will require 5*0.2 = 1.0mol O2 </span>
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4 0
3 years ago
What is a solubility?
kogti [31]

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Explanation:

6 0
2 years ago
purification of chromium can be achieved by electrorefining chromium from an impure chromium anode onto a pure chromium cathode
stiv31 [10]

The chromium in the electrolytic solution is present as cr³⁺ will take 6.19 hours  will it take to plate 18.0 kg of chromium onto the cathode if the current passed through the cell is held constant at 45.0 A.

It is given that mass is 18 kg. Converting mass into grams

18 x (1000g / 1 kg)

= 18000 g

So, 18 kg = 18000 g

Now, calculate the number of moles

No. of moles = mass of Cr / molecular mass of Cr

No. of moles = 18000 / 52 g/mole

No. of moles = 346.15 mole

For , Cr³⁺, 3 moles of electrons are required

Hence,

3 x 346.15 mole

1038.45

As 96500 C of charge are present in 1 mole of electrons. As a result, the charge held by 1038.45 mole of electrons will be determined as follows.

= 1038.45mol x  96500 C / mole

= 10.02 x 10⁷ C

As, we know that relation between charge, current and time is as follows.

Q = I x T

Current is given as 45 A

Therefore, charge is calculated  

T = Q / I

T = 100.21 x 10⁶ / 45

T = 2.227 x 10⁶

As there are 3600 seconds in one 1 hour. Converting 2.227 x 10⁶ into hours are

= 2.227 x 10⁶ / 3600 sec/h

= 6.19 hours

Thus, if the current flowing through the cell is maintained constant at 45 A, we may estimate that it will take 6.19 hours to plate 18 kg of chromium onto the cathode.

To know more about  electrolytic solution visit :

brainly.com/question/9203208

#SPJ4

4 0
1 year ago
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