Titanium reacts with bromine gas according to the following balanced chemical equation:
2Ti(s)+3Br2(l)→2TiBr3(s)
A: What mass of titanium(III) bromide is produced when 139.29 g titanium reacts with 553.73 g bromine liquid? in gTiBr3
B: What is the percent yield of titanium(III) bromide if 556.03 g of titanium(III) bromide is actually obtained?
A)
Molar mass of Ti = 47.87 g/mol
mass(Ti)= 139.29 g
use:
number of mol of Ti,
n = mass of Ti/molar mass of Ti
=(1.393*10^2 g)/(47.87 g/mol)
= 2.91 mol
Molar mass of Br2 = 159.8 g/mol
mass(Br2)= 553.73 g
use:
number of mol of Br2,
n = mass of Br2/molar mass of Br2
=(5.5373*10^2 g)/(1.598*10^2 g/mol)
= 3.4651 mol
Balanced chemical equation is:
2 Ti + 3 Br2 ---> 2 TiBr3
2 mol of Ti reacts with 3 mol of Br2
for 2.91 mol of Ti, 4.365 mol of Br2 is required
But we have 3.4651 mol of Br2
so, Br2 is limiting reagent
we will use Br2 in further calculation
Molar mass of TiBr3,
MM = 1*MM(Ti) + 3*MM(Br)
= 1*47.87 + 3*79.9
= 287.57 g/mol
According to balanced equation
mol of TiBr3 formed = (2/3)* moles of Br2
= (2/3)*3.4651
= 2.3101 mol
use:
mass of TiBr3 = number of mol * molar mass
= 2.3101* 287.57
= 664.31 g
Answer: 664.31 g
B)
% yield = actual mass*100/theoretical mass
= 556.03*100/664.31
= 83.700 %
Answer: 83.700 %
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