d 1.) 3.14 grams of silver nitrate are mixed with excess iron(II) chloride . The

Question

d1.) 3.14 grams of silver nitrate are mixed with excess iron(II) chloride. The reaction yields 1.40 grams of iron(II) nitrate.

What is the theoretical yield of iron(II) nitrate ? grams

What is the percent yield of iron(II) nitrate ? %

2.)5.65 grams of nitrogen gas are mixed with excess oxygen gas. The reaction yields 8.87 grams of nitrogen monoxide.

What is the theoretical yield of nitrogen monoxide ?

What is the percent yield for this reaction ?

3.) 5.81 grams of silver nitrate are mixed with excess copper. The reaction yields 2.54 grams of copper(II) nitrate.

What is the theoretical yield of copper(II) nitrate ?

What is the percent yield of copper(II) nitrate ?

Explanation / Answer

(1) FeCl2 + 2 AgNO3 => Fe(NO3)2 + 2 AgCl(s)

Moles of AgNO3 = mass/molar mass of AgNO3

= 3.14/169.87 = 0.018485 mol

Moles of Fe(NO3)2 = 1/2 x moles of AgNO3

= 1/2 x 0.018485 = 0.009242 mol

Theoretical yield = moles x molar mass of Fe(NO3)2

= 0.009242 x 179.85

= 1.662 g = 1.66 g

Percent yield = actual yield/theoretical yield x 100%

= 1.40/1.662 x 100% = 84.2%

(2) N2(g) + O2(g) => 2 NO(g)

Moles of N2 = mass/molar mass of N2

= 5.65/28.01 = 0.2017 mol

Moles of NO = 2 x moles of N2

= 2 x 0.2017 = 0.4034 mol

Theoretical yield = moles x molar mass of NO

= 0.4034 x 30.006

= 12.11 g = 12.1 g

Percent yield = actual yield/theoretical yield x 100%

= 8.87/12.11 x 100% = 73.3%

(3) 2 Ag(NO3) + Cu => Cu(NO3)2 + 2 Ag

Moles of AgNO3 = mass/molar mass of AgNO3

= 5.81/169.87 = 0.03420 mol

Moles of Cu(NO3)2 = 1/2 x moles of AgNO3

= 1/2 x 0.03420 = 0.01710 mol

Theoretical yield = moles x molar mass of Cu(NO3)2

= 0.01710 x 187.56

= 3.208 g = 3.21 g

Percent yield = actual yield/theoretical yield x 100%

= 2.54/3.208 x 100% = 79.2%

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