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laila [671]
3 years ago
15

In a chemical reaction, substance R reacts with compound XY to produce 47.0 g of compound RX. If the theoretical yield of RX is

56.0 g, what is its percent yield?
A.9.00%
B.16.1%
C.83.9%
D.119%
Chemistry
2 answers:
RideAnS [48]3 years ago
6 0

Answer: The correct option is C.

Explanation: For a given chemical reaction:

R+XY\rightarrow RX+Y

The mass of product, RX formed in the reaction = 47.0 g

Theoretical yield of the product, RX = 56.0 g

To calculate percent yield, we use the formula:

\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

Putting the values in above equation, we get:

\%\text{ yield}=\frac{47.0g}{56.0g}\times 100

Percent yield = 83.9%

sdas [7]3 years ago
3 0
The answer is C 83.9%.
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Answer:

Approximately 1.876 \times 10^{-3}\; \rm mol.

Explanation:

Convert both volumes to standard units (that is: liters.)

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n(\mathrm{HCl}, \, \text{initial}) = \displaystyle c \cdot V = 1.00\times 1.000 \times 10^{-2}= 1.000\times 10^{-2}\; \rm mol.

Number of moles of \rm NaOH from the titration:

n(\mathrm{NaOH}) = c \cdot V = 0.30 \times 2.708 \times 10^{-2} = 8.124 \times 10^{-3}\; \rm mol.

\rm NaOH neutralizes \rm HCl at a 1:1 ratio:

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\begin{aligned}&n(\mathrm{HCl},\, \text{consumed}) \\ =& n(\mathrm{HCl},\, \text{initial}) - n(\mathrm{HCl},\, \text{leftover}) \\ =& 1.000\times 10^{-2} - 8.124\times 10^{-3} \\ =& 1.876 \times 10^{-3}\; \rm mol\end{aligned}.

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