Question

for the following reaction, 5.19 grams of iron(ii) chloride are mixed with excess silver nitrate. the reaction yields 6.00 grams of iron(ii) nitrate.
iron(ii) chloride (aq) + silver nitrate (aq) → iron(ii) nitrate (aq) + silver chloride (s)
what is the theoretical yield of iron(ii) nitrate? grams
what is the percent yield of iron(ii) nitrate? %
submit answer retry entire group 2 more group attempts remaining
the equation for this reaction is:
fecl₂ (aq) + 2agno₃ (aq) → fe(no₃)₂ (aq) + 2agcl (s)

Explanation:

Step1: Calculate moles of \(FeCl_2\)

Molar mass of \(FeCl_2 = 55.85+(2\times35.45)=126.75\space g/mol\)
Moles of \(FeCl_2=\frac{5.19\space g}{126.75\space g/mol}\approx0.041\space mol\)

Step2: Determine moles of \(Fe(NO_3)_2\) from mole ratio

From the balanced equation \(FeCl_2(aq)+2AgNO_3(aq)\to Fe(NO_3)_2(aq) + 2AgCl(s)\), mole ratio \(FeCl_2:Fe(NO_3)_2 = 1:1\)
So moles of \(Fe(NO_3)_2=0.041\space mol\)

Step3: Calculate theoretical yield of \(Fe(NO_3)_2\)

Molar mass of \(Fe(NO_3)_2=55.85+(2\times14)+(6\times16)=179.85\space g/mol\)
Theoretical yield \(=0.041\space mol\times179.85\space g/mol\approx7.37\space g\)

Step4: Calculate percent yield

Percent yield \(=\frac{actual\space yield}{theoretical\space yield}\times100\%=\frac{6.00\space g}{7.37\space g}\times100\%\approx81.4\%\)

Answer:

Theoretical yield: \(7.37\) grams
Percent yield: \(81.4\%\)