Q160CP

Question

When powdered zinc is heated with sulfur, a violent reaction occurs, and zinc sulfide forms:

$Z n (s) + S_{8} (s) \to Z n S (s) [u n b a l a n c e d]$

Some of the reactants also combine with oxygen in air to form zinc oxide and sulfur dioxide. When 83.2 g of Zn reacts with 52.4 g of S₈, 104.4 g of ZnS forms. What is the percent yield of ZnS? (b) If all the remaining reactants combine with oxygen, how many grams of each of the two oxides form?

Step-by-Step Solution

Verified

Answer

The percent yield is 84.3%.
There are 16.4 g ZnO and 35.8 g SO₂ produced

1Step 1: Finding the percent yield

The moles of ZnS can be found as,

$M o l e s o f Z n S (f r o m Z n) = 83.2 g Z n \times \frac{1 m o l Z n}{65.38 g Z n} \times \frac{8 m o l Z n S}{8 m o l Z n} = 1.27 m o l Z n S M o l e s o f Z n S (f r o m S_{8}) = 52.4 g S_{8} \times \frac{1 m o l S_{8}}{256.56 g S_{8}} \times \frac{8 m o l Z n S}{1 m o l S_{8}} = 1.63 m o l Z n S$

2Step 2: Calculate mass of zinc and its percent yield:

The mass of Zn and percent yield can be denoted as,

$M a s s o f Z n (t h e o r e t i c a l) = 1.27 m o l Z n S \times \frac{97.47 g Z n S}{1 m o l Z n S} = 123.8 g Z n S % y i e l d = \frac{a c t u a l y i e l d}{t h e o r e t i c a l y i e l d} \times 100 % y i e l d = \frac{104.4 g}{123.8 g} \times 100 % y i e l d = 84.3 %$

3Step 2: Finding the production of each of the two oxides

The mass for Zn and S₈ can be formed as,

$M a s s o f Z n (f o r Z n S) = 104.4 g Z n S \times \frac{1 m o l Z n S}{97.47 g Z n S} \times \frac{8 m o l Z n}{8 m o l Z n S} \times \frac{65.38 g Z n}{1 m o l Z n} = 70 g Z n M a s s o f S_{8} (f o r Z n S) = 104.4 g Z n S \times \frac{1 m o l Z n S}{97.47 g Z n S} \times \frac{1 m o l S_{8}}{8 m o l Z n S} \times \frac{256.56 g S_{8}}{1 m o l S_{8}} = 34.5 g S_{8}$

On subtracting these masses,

$M a s s o f Z n (f r o m Z n O) = 83.2 g - 70 g = 13.2 g Z n M a s s o f S_{8} (f o r S O_{2}) = 52.4 g - 34.5 g = 17.9 S_{8}$

The mass of produced oxides as follows,

$M a s s o f Z n O = 13.2 g Z n \times \frac{1 m o l Z n}{65.38 g Z n} \times \frac{2 m o l Z n O}{2 m o l Z n} \times \frac{81.38 g Z n O}{1 m o l Z n O} = 16.4 g Z n O M a s s o f S O_{2} = 17.9 g S_{8} \times \frac{1 m o l S_{8}}{256.56 g S_{8}} \times \frac{8 m o l S O}{1 m o l S_{8}} \times \frac{64.06 g S O_{2}}{1 m o l S O_{2}} = 35.8 g S O_{2}$

Q159CP

Q161CP

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