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3 years ago

What is the percentage yield if 125.4g C3H8 are collected from a reaction that should produce 157.4g C3h8?

Chemistry

1 answer:

cestrela7 [59]3 years ago

3 0

The percentage yield : 79.67%

The closest answer is option B.

<h3>Further explanation</h3>

Given

125.4 g C3H8

157.4 g C3H8

Required

The percentage yield

Solution

Percent yield is the comparison of the amount of product obtained from a reaction with the amount you calculated

General formula:

% yield = (actual yield : theoretical yield) x 100%

Input the value :

% yield = (125.4 : 157.4) x 100%

% yield = 79.67

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Analysis of a compound of sulfur, oxygen and fluorine showed that it is 31.42% S and 31.35% O, with F accounting for the remaind

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Answer:

The molecular formula is SO2F2

Explanation:

Step 1: Data given

Suppose the mass of compound = 100 grams

The compound contains:

31.42 % S = 31.42 grams S

31.35 % O = 31.35 grams O

100 - 31.42 - 31.35 = 37.23 F

Molar mass of S = 32.065 g/mol

Molar mass F = 19.00 g/mol

Molar mass O = 16.00 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles S = 31.42 grams / 32.065 g/mol

Moles S = 0.9799 moles

Moles 0 = 31.35 grams / 16.00 g/mol

Moles 0 = 1.959 moles

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Moles F = 1.959 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

S: 0.9799 / 0.9799 = 1

F: 1.959/ 0.9799 = 2

O : 1.959 / 0.9799 = 2

The empirical formula is SO2F2

This formula has a molecular mass of 102.06 g/mol

This means the empirical formula is also the molecular formula : SO2F2

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