Question

If 3.00 g of titanium metal is reacted with 6.00 g of chlorine gas, Cl2, to form 7.7 g of titanium (IV) chloride in a combination reaction, what is the percent yield of the product?

Answer 1

Ti + 2 Cl2 → TiCl4  

(3.00 g Ti) / (47.867 g Ti/mol) = 0.062674 mol Ti  

(6.00 g Cl2) / (70.9064 g Cl2/mol) = 0.084619 mol Cl2  

0.084619 mole of Cl2 would react completely with 0.084619 x (1/2) = 0.0423095 mole of Ti, but there is more Ti present than that, so Ti is in excess and Cl2 is the limiting reactant.  

(0.084619 mol Cl2) x (1 mol TiCl4 / 2 mol Cl2) x (189.679 g TiCl4/mol) = 8.025 g TiCl4 in theory  

(7.7 g) / (8.025 g) = 0.96 = 96% yield TiCl4

Answer 2

Answer: The percent yield of titanium chloride is 96.73 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

• For titanium:

Given mass of titanium = 3 g

Molar mass of titanium = 47.87 g/mol

Putting values in equation 1, we get:

$\text{Moles of titanium}=\frac{3g}{47.87g/mol}=0.063mol$

• For chlorine gas:

Given mass of chlorine gas = 6 g

Molar mass of chlorine gas = 71 g/mol

Putting values in equation 1, we get:

$\text{Moles of chlorine gas}=\frac{6g}{71g/mol}=0.084mol$

• The chemical equation for the reaction of titanium and chlorine gas follows:

$Ti+2Cl_2\rightarrow TiCl_4$

By Stoichiometry of the reaction:

2 moles of chlorine gas reacts with 1 mole of titanium metal.

So, 0.084 moles of chlorine gas will react with = $\frac{1}{2}\times 0.084=0.042mol$ of titanium metal.

As, given amount of titanium metal is more than the required amount. So, it is considered as an excess reagent.

Thus, chlorine gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of chlorine gas produces 1 mole of titanium (IV) chloride.

So, 0.084 moles of chlorine gas will produce = $\frac{1}{2}\times 0.084=0.042moles$ of titanium (IV) chloride.

• Now, calculating the mass of titanium (IV) chloride from equation 1, we get:

Molar mass of titanium (IV) chloride = 189.68 g/mol

Moles of titanium (IV) chloride = 0.042 moles

Putting values in equation 1, we get:

$0.042mol=\frac{\text{Mass of titanium (IV) chloride}}{189.68g/mol}\\\\\text{Mass of titanium (IV) chloride}=7.96g$

• To calculate the percentage yield of titanium (IV) chloride, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of titanium (IV) chloride = 7.7 g

Theoretical yield of titanium (IV) chloride = 7.96 g

Putting values in above equation, we get:

$\%\text{ yield of titanium (IV) chloride}=\frac{7.7g}{7.96g}\times 100\\\\\% \text{yield of titanium (IV) chloride}=96.73\%$

Hence, the percent yield of the reaction is 96.73 %.