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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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3 years ago
How many moles of oxygen are formed when 58.6 g of KNO3 decomposes according to the following reaction? 4 KNO3(s) → 2 K2O(s) + 2
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Answer:

0.725 mol

Explanation:

Moles are calculated as the given mass divided by the molecular mass.

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moles of KNO₃ = 58.6 g / 101 g / mol

moles of KNO₃ = 0.58 mol

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