In a certain experiment, 5.26 g of Fe(OH), (formula mass 107) was produced from
ID: 540182 • Letter: I
Question
In a certain experiment, 5.26 g of Fe(OH), (formula mass 107) was produced from 12.5 g of Fe(NO,), (formula mass 242) reacting with an excess of NH, and H,O according to the equation Fe(NOJ, + 3NH, + 3H Fe(OH), + 3NH.NO3 What is the percent yield of the reaction? a) b) c) 0 31.1% 83.7% 69.8% d) e) 95.1% 42.1% In a certain experiment using Zn and an excess of S, 29.9 g of ZnS is obtained. This represents a 89.4 percent yield. What is the theoretical yield of ZnS for this experiment? a) 29.2 g b) 30.6g c) 33.4 g d) 35.1g e) 27.3 g The following two-step process can be used to produce So2: 3) If 8.00 moles of S02 were produced, how many moles of KC103 were de a) 4.33 moles b) 6.66 moles c) 8.00 moles d) 16.00 moles e) 5.33 moles sed to initiate the sequence of reactions?Explanation / Answer
1)
Molar mass of Fe(NO3)3 = 242 g/mol
mass of Fe(NO3)3 = 12.5 g
molar mass of Fe(NO3)3 = 241.88 g/mol
mol of Fe(NO3)3 = (mass)/(molar mass)
= 12.5/242
= 0.0517 mol
According to balanced equation
mol of Fe(OH)3 formed = moles of Fe(NO3)3
= 0.052 mol
mass of Fe(OH)3 = number of mol * molar mass
= 0.0517*107
= 5.53 g
% yield = actual mass*100/theoretical mass
= 5.26*100/5.53
= 95.1 %
Answer: d
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