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Temka [501]
3 years ago
7

A reaction has a theoretical yield of 124.3 g SF6, but only 113.7 g SF6 are obtained in the lab, what is the percent yield of SF

6 for this reaction?
Chemistry
1 answer:
ratelena [41]3 years ago
6 0

Answer:

Percent yield = 91%

Explanation:

Given data:

Theoretical yield of SF₆ = 124.3 G

Actual yield of SF₆ = 113.7 g

Percent yield of SF₆ = ?

Solution:

Formula:

Percent yield = (actual yield / theoretical yield)× 100

By putting values,

Percent yield = (113.7 g/ 124.3 g) × 100

Percent yield = 0.91 × 100

Percent yield = 91%

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Answer:

      \large\boxed{\large\boxed{CH_6N_2}}

Explanation:

<u><em>1. First determine the empirical formula.</em></u>

a) Base: 100 g of compound

             mass       atomic mass      number of moles

                g                g/mol                    mol

C           26.06            12.011             26.06/12.011   = 2.17

H           13.13              1.008              13.13/1.008    =  13.03

N           60.81             14.007            60.81/14.007 = 4.34

b) Divide every number of moles by the smallest number: 2.17

mass       number of moles        proportion

C                  2.17/2.17                         1

H               13.03/2.17                         6

N                4.34/2.17                          2

c) Empirical formula

      CH_6N_2

d) Mass of the empirical formula

      1\times 12.011g/mol6+6\times 1.008g/mol+2\times 14.007g/mol=46.07g/mol

<u><em>2. Molecular formula</em></u>

Since the mass of one unit of the empirical formula is equal to the molar mass of the compound, the molecular formula is the same as the empirical formula:

                   CH_6N_2

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