Question #6e6ac

2 Answers
Sep 10, 2017

Answer:

We need a stoichiometrically balanced chemical equation to represent the redox reaction.....and I gets approx. 75% yield.

Explanation:

The metal is oxidized, and the halogen is reduced.....

#Ti(s) + 2Cl_2(g) rarr TiCl_4(l)#

And so we interrogate the molar quantities of the reactants....

#"Moles of metal"# #=# #(3.0*g)/(47.87*g*mol^-1)=0.0627*mol#.

#"Moles of halogen"# #=# #(6.0*g)/(70.90*g*mol^-1)=0.1086*mol#, i.e. #0.1086*mol# with respect to #Cl_2(g)#

#"Moles of titanic chloride"# #=# #(7.7*g)/(189.68*g*mol^-1)=0.0406*mol#.

Given the 1:2 stoichiometry inherent in the balanced equation, CLEARLY, the halogen is the limiting reagent. And thus AT MOST #(0.1086*mol)/2# of #"titanic chloride"# can be produced....

And thus (finally)....

#"% yield"=(0.0406*mol)/(1/2xx0.1086*mol)xx100%=??%#

Note that in this scenario, i.e. limited quantities of the halogen, incomplete oxidation to #TiCl_2#, and #TiCl_3# may occur....

Sep 10, 2017

Answer:

WARNING! Long answer! The percent yield is 96 %.

Explanation:

We are given the masses of two reactants, so this is a limiting reactant problem.

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and everything else below them.

We start with the balanced equation.

#M_r:color(white)(mmml) 47.87color(white)(mmm)70.91color(white)(mml)189.68#
#color(white)(mmmmmml) "Ti"color(white)(m) +color(white)(ml) "2Cl"_2 → color(white)(m)"TiCl"_4#
#"Mass/g:"color(white)(mml)3.0color(white)(mmll)6.00color(white)(mmmml)7.76#
#"Moles:" color(white)(mmm)0.0627color(white)(m)"0.084 61"color(white)(mll)#
#"Divide by:"color(white)(mml)1color(white)(mmmm)2#
#"Moles rxn:"color(white)(m)0.0626color(white)(m)"0.042 31"#

(a) Calculate the moles of #"Ti"#

#"Moles of Ti" = 3.0 color(red)(cancel(color(black)("g Ti"))) × ("1 mol Ti")/(47.87 color(red)(cancel(color(black)("g Ti")))) = "0.0626 mol Ti"#

(b) Calculate moles of #"Cl"_2#

#6.00color(red)(cancel(color(black)("g Cl"_2))) × ("1 mol Cl"_2)/(70.91color(red)(cancel(color(black)("g Cl"_2)))) = "0.084 61 mol Cl"_2#

2. Identify the limiting reactant

An easy way to identify the limiting reactant is to calculate the "moles of reaction" each will give:

You divide the moles of each reactant by its corresponding coefficient in the balanced equation.

I did that for you in the table above.

#"Cl"_2# is the limiting reactant because it gives the fewer moles of reaction.

3. Calculate the theoretical yield of #"TlCl"_4#.

#"Theoretical yield" = "0.084 61" color(red)(cancel(color(black)("mol Cl"_2))) × (1 color(red)(cancel(color(black)("mol TlCl"_4))))/(2 color(red)(cancel(color(black)("mol Cl"_2)))) × "189.68 g TlCl"_4/(1 color(red)(cancel(color(black)("mol TlCl"_4)))) = "8.025 g TlCl"_4#

The theoretical yield of #"TlCl"_4# is 8.025 g.

4. Calculate the percent yield of #"TiCl"_4#

The formula for percentage yield is

#color(blue)(bar(ul(|color(white)(a/a)"% yield" = "actual yield"/"theoretical yield" × 100 %color(white)(a/a)|)))" "#

#"Percent yield" = (7.7 color(red)(cancel(color(black)("g"))))/(8.024 color(red)(cancel(color(black)("g")))) × 100 % = 96 %#