Consider the following reaction: 4 Fe (s) + 3 O_2 (g) rightarrow 2 Fe_2O_3 (s) T

Question

Consider the following reaction: 4 Fe (s) + 3 O_2 (g) rightarrow 2 Fe_2O_3 (s) The formula mass for Fe is g/mol. a. 223.4 b. 55.85 c. 111.7 d 6.022 times 10^23 The formula mass for Fe_2O_3 is g/mol. a. 319.4 b. 73 85 c. 199.6 d. 159.7 The molar ratio of Fe to Fe_2O_3 is a. 1:1 b. 3:2 c. 4:2 d 2:2 How many grams of Fe_2O_3, will be produced when 5.00 g Fe reacts in oxygen? a. 6.53 g b. 0.957 g c 2 50 g d 70.5 g when a student tried this experiment, she recovered 0.884 g Fe_2O_3 What is her percent yield? a 1.08% b. 0.924% c. 108% d. 92 4%

Explanation / Answer

Part A,B and C are already answered.

Part-D:-

From the given balance equation we can say that 4 mole of Fe(molar mass=55.85g) produced 2 mole Fe2O3(molar mass=159.7g)

We can also say 223.4g Fe produced 319.4g Fe2O3. So 5g Fe would produce approx 6.5g Fe2O3.

Part-E:-

% yield=(actual/theoretical)*100=(0.684/6.53)*100=10.8%

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