Question

a)

For the following reaction, 5.52 grams of oxygen gas are mixed with excess iron . The reaction yields 21.0 grams of iron(II) oxide .

iron ( s ) + oxygen ( g ) iron(II) oxide ( s )

What is the theoretical yield of iron(II) oxide ?	grams
What is the percent yield for this reaction ?	%

b) For the following reaction, 6.29 grams of hydrogen gas are mixed with excess iodine . The reaction yields 745 grams of hydrogen iodide .

hydrogen ( g ) + iodine ( s ) hydrogen iodide ( g )

What is the theoretical yield of hydrogen iodide ?	grams
What is the percent yield for this reaction ?	%

Answer 1

a)

i)

Molar mass of O2 = 32 g/mol

mass(O2)= 5.52 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(5.52 g)/(32 g/mol)

= 0.1725 mol

Balanced chemical equation is:

O2 + 2 Fe ---> 2 FeO

Molar mass of FeO,

MM = 1*MM(Fe) + 1*MM(O)

= 1*55.85 + 1*16.0

= 71.85 g/mol

According to balanced equation

mol of FeO formed = (2/1)* moles of O2

= (2/1)*0.1725

= 0.345 mol

use:

mass of FeO = number of mol * molar mass

= 0.345*71.85

= 24.79 g

Answer: 24.8 g

ii)

% yield = actual mass*100/theoretical mass

= 21*100/24.8

= 84.7 %

Answer: 84.7 %

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