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kenny6666 [7]
4 years ago
15

Will mark branliest. Calculating percent yield: For the reaction Cu + 2AgNO3 → Cu(NO3)2 + 2 Ag, adding 127 grams of copper to lo

ts of silver nitrate yields 410 grams of silver. What is the percent yield of silver?
I'm desperate.
Chemistry
1 answer:
pishuonlain [190]4 years ago
8 0

Percent yield = (Actual Yield/ Theoretical Yield) x 100

Given, Actual Yield of silver = 410 grams

Given, mass of copper = 127 g

Atomic mass of copper = 63.546 amu

Formula: Moles = mass / atomic mass

Moles of copper = 127 g/ 63.546 amu = 1.998 mol

Based on the given balanced chemical reaction, the molar ratio between Cu: Ag is 1:2

So 1.998 mol of copper should yield (2 mol Ag/ 1 mol Cu) x 1.998 mol of Cu = 3.996 mol

Calculated mol of Ag = 3.996 mol

Atomic mass of silver = 107.8682 amu

Mass of silver = moles x atomic mass = 3.996 mol x 107.8682 amu = 431 g

Based on the math, the theoretical yield = 431 g

Percent yield of silver = (410g/431g) x 100 = 95. 13%

The answer is 95. 13%

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A system does 125 J of work and cools down by releasing 438 J of heat. The change in internal energy is ____________ J.
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Given that,

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A reaction produced 37.5 L of oxygen gas at 307 K and 1.25 atm. How many moles of oxygen were produced? 0.538 mol, O2 1.86 mol,
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Answer:

O 2  has a mass of 1.78g

Explanation:

We are at STP and that means we have to use the ideal gas law equation!

P represents pressure (could have units of atm, depending on the units of the universal gas constant)

V represents volume (must have units of liters)

n represents the number of moles

R is the universal gas constant (has units of  

L

× a t m

m o l

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)

T represents the temperature, which must be in Kelvins.

Next, list your known and unknown variables. Our only unknown is the number of moles of  

O

2

(

g

)

. Our known variables are P,V,R, and T.

At STP, the temperature is 273K and the pressure is 1 atm. The proportionality constant, R, is equal to 0.0821  

L

×

a

t

m

m

o

l

×

K

Now we have to rearrange the equation to solve for n

n

=

P

V

R

T

n

=

1

atm

×

1.25

L

0.0821

Lxxatm

m

o

l

×

K

×

273

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n

=

0.05577

m

o

l

To get the mass of  

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0.0577

mol

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2

×

32.00

g

1

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