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ANEK [815]
2 years ago
13

A reaction produces 10.5 L if oxygen, but was supposed to produce 1 mol of oxygen. What is the percent yield ? (One mole of any

gas occupies 22.4 L under certain conditions of temperature and pressure. Assume those conditions for this question)
Chemistry
1 answer:
Rzqust [24]2 years ago
5 0
To answer this item, we assume that oxygen behaves ideally such that it is able to fulfill the following equation,

 PV = nRT

If we are to retain constant the variable n and V. 

The percent yield can therefore be solved through the following calculation,

   n = (10.5 L)/(22.4 L)   x 100%

Simplifying,
    n = 46.875%

Answer: 48.87%
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An unknown compound with a molar mass of 223.94 g/mol consists of 32.18% c, 4.50% h, and 63.32% cl. find the molecular formula f
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The actual number of atoms of each element present in the molecule of the compound is represented by the formula known as molecular formula.

Molar mass of the unknown compound = 223.94 g/mol (given)

Mass of each element present in the unknown compound is determined as:

  • Mass of carbon, C:

\frac{32.18}{100}\times 223.94 = 72.06 g

  • Mass of hydrogen, H:

\frac{4.5}{100}\times 223.94 = 10.08 g

  • Mass of chlorine, Cl:

\frac{63.32}{100}\times 223.94 = 141.79 g

Now, the number of each element in the unknown compound is determined by the formula:

number of moles = \frac{given mass}{molar mass}

  • Number of moles of C:

number of moles = \frac{72.06}{12} = 6.005 mole\simeq 6 mole

  • Number of moles of H:

number of moles = \frac{10.08}{1} = 10.08 mole\simeq 10 mole

  • Number of moles of Cl

number of moles = \frac{141.79}{35.5} = 3.99 mole\simeq 4 mole

Dividing each mole with the smallest number of mole, to determine the empirical formula:

C_{\frac{6}{4}}H_{\frac{10}{4}}Cl_{\frac{10}{4}}

C_{1.5}H_{2.5}Cl_{1}

Multiplying with 2 to convert the numbers in formula into a whole number:

So, the empirical formula is C_{3}H_{5}Cl_{2}.

Empirical mass = 12\times 3+1\times 5+2\times 35.5 = 112 g/mol

In order to determine the molecular formula:

n = \frac{molar mass}{empirical mass}

n = \frac{223.94}{112} = 1.99 \simeq 2

So, the molecular formula is:

2\times C_{3}H_{5}Cl_{2} =  C_{6}H_{10}Cl_{4}

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