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Tresset [83]
3 years ago
14

CuCl2 + 2NaNO3 mc023-1.jpg Cu(NO3)2 + 2NaCl What is the percent yield of NaCl if 31.0 g of CuCl2 reacts with excess NaNO3 to pro

duce 21.2 g of NaCl?
Chemistry
2 answers:
MatroZZZ [7]3 years ago
3 0
The stoichiometric ratio of CuCl2 to NaCl is 1 is to 2. The stoichiometric ratio of 31.0 g CuCl2 is 26.95 grams of NaCl by converting the amount of CuCl2 to mole and multiplying by 0.5 and molar mass of NaCl.This amount is equal to 78.65% yield.
Lera25 [3.4K]3 years ago
3 0
The answer is D 78.7%
just finished the test
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If 495 milliliters of carbon dioxide at 25°C and 101.3 kilopascals reacts with excess water, what is the theoretical yield of ca
postnew [5]

Answer:- 1.24 g

Solution:- The balanced equation for the formation of carbonic acid by the reaction of carbon dioxide with water is:

CO_2+H_2O\rightarrow H_2CO_3

From balanced equation, there is 1:1 mol ratio between carbon dioxide and carbonic acid. So, moles of carbonic acid will be equal to the moles of carbon dioxide used.

Moles of carbon dioxide can be calculated using ideal gas law equation as it's volume, temperature and pressure are given.

we need to convert mL to L, degree C to kelvin and kilopascals to atm.

495mL(\frac{1L}{1000mL})  = 0.495 L

25 + 273 = 298 K

101.3kPa(\frac{1atm}{101.3kPa)})

= 1 atm

Ideal gas law equation is:

PV = nRT

We want to find out the n, so let's rearrange this:

n=\frac{PV}{RT}

R is the universal gas constant and it's value is \frac{0.0821atm.L}{mol.K} .

Let's pug in the values in the equation and solve it for n.

n=\frac{1*0.495}{0.0821*298}

n = 0.02 mol

To convert the moles to grams we multiply the moles by the molar mass of carbonic acid.

Molar mass of carbonic acid = 2(1.008)+12.01+3(16.00)

= 2.016+12.01+48.00

= 62.03 gram per mol (rounded to two decimal places)

Let's multiply the moles by molar mass:

0.02mol(\frac{62.03g}{mol})

= 1.24 g

So, the theoretical yield of carbonic acid is 1.24 g.


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