Question

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Question:a

Titanium reacts with chlorine gas according to the following balanced chemical equation: 2Ti(s)+3Cl2(g)-->2TiCl3(s)

A.) What mass of Titanium (III) Chloride is produced when 129.72 g titanium reacts with 281.14 g chlorine gas?

B.) What is the percent yield of titanium (III) chloride if 259.69 g of titanium (III) chloride is actually obtained?

Answer 1

Answer:

Explanation:

A mole ratio is ​the ratio between the amounts in moles of any two compounds involved in a chemical reaction. The mole ratio can be determined by examining the coefficients in front of formulas in a balanced chemical equation.

Step 1: write the balanced chemical equation.

2Ti(s)+3Cl₂(g)------>2TiCl₃(s)

Step 2: calculate the moles of Ti and Cl₂

Moles of Ti = mass given / molar mass = ( 129.72 g / 47.867 g/mol ) = 2.71 mol

Moles of Cl₂= (281.14 g / 70.906 g/mol ) = 3.965 mol

Step 3: Determine the limiting reagent

2Ti(s)+3Cl₂(g)------>2TiCl₃(s)

According to the reaction we need
3 mol of Cl₂ requires for 2 mol of Ti
so, for 3.965 mol of Cl₂ we will need =(2 mol of Ti / 3 mol of Cl₂ )× 3.965 mol of Cl₂ = 2.64 mol of Ti
Since we need only 2.64 mol of Ti hence Ti is in excess ( since given = 2.71 mol ) and Cl₂ limiting reagent.

so the amount of product will formed according to the limiting reagent ( i.e Cl₂ amount )

Step 3: Calculate the moles of TiCl₃ produced

2Ti(s)+3Cl₂(g)------>2TiCl₃(s)

According to the reaction:
3 mol of Cl₂ produce 2 mol of TiCl₃
so, 3.965  mol of Cl₂ will produce=( 2 mol of TiCl₃ / 3 mol of Cl₂ )× 3.965 mol of Cl₂ = 2.6433 mol of TiCl₃

(A) Step 4: Calculation of TiCl₃produced

we get moles of TiCl₃ that can be produced = 2.6433  mol

Mass of TiCl₃ produced =( moles×molar mass ) = ( 2.6433 mol × 154.225 g/mol ) = 407.66 g

(B) step 5: Calculation of percent yield of titanium (III) chloride

Given, actual yield = 259.69 g

We got theoretical yield = 407.66 g

We know, Percent yield = (actual yield / theoretical yield ) × 100 %

Percent yield = (259.69 / 407.6647627 ) × 100 % = 63.7 %