Q5. For the reaction shown, find the limiting reactant for each of the following

Question

Q5. For the reaction shown, find the limiting reactant for each of the following initial amounts of reactants.
2Na(s)+Br2(g)2NaBr(s)

A )

4 mol Na, 4 mol Br2

Express your answer as a chemical formula.

B)

1.8 mol Na, 1.4 mol Br2

Express your answer as a chemical formula.

Part C

2.5 mol Na, 1 mol Br2

Express your answer as a chemical formula.

Part D

12.0 mol Na, 6.9 mol Br2

Express your answer as a chemical formula.

= = = = =

Q 6 .

Consider the reaction: ____Al + ____Fe2O3   ____Al2O3 + ____Fe

Part A

Balance the equation.

Write the coefficients from left to right, separated by commas. For example: 1,2,3

Part B

If you start with 53g of Al and 123g of Fe2O3, what is your limiting reactant?

Part C

How many grams of Fe can you make?

Part D

If you do the experiment in lab and produce 63.2 g of Fe, what is your percent yield?

Explanation / Answer

reactants.
2Na(s)+Br2(g)2NaBr(s)

A )

4 mol Na, 4 mol Br2

As per the stoichiometry of the reaction, 2 moles of Na requires 1 mole of Br2 and 4 moles of Na requires 2 moles of Br2. But Br2 is 4 moles and hence limiting reactants is Na and excess reactants is Br2.

Express your answer as a chemical formula.

B)

1.8 mol Na, 1.4 mol Br2

Molar ratio as per stoichiometry of Na : Br2 =2 :1

Given ratio = 1.8 :1.4 =   1.8/1.4: 1 = 1.3 :1

So sodium is limiting reactant and Br2 is excess reactant

Express your answer as a chemical formula.

Part C

2.5 mol Na, 1 mol Br2

Given ratio = 2.5:1 stoichiometric ratio = 2:1

So excess is Na and limiting is Br2.

Part D

12.0 mol Na, 6.9 mol Br2

Given ratio = 12 :6.9 =12/6.9 :1 =1.7 :1

Excess is Br2 and limiting is Na

2.

The coefficients are 2,1,1,2

Moles of Al in 53 gms = 53/27=1.96 and moles of Fe2O3= 123/ 160= 0.8

Stoichiometric requirements = 2:1

Actual requirement = 1.96 :0.8 =2.45 :1

Excess reactant is Al and limiting reactant is Fe2O3.

Part C

1 mole of Fe2O3 gives 2 moles of Fe

0.8 moles of Fe2O3 gives 0.8*2= 1.6 moles of Fe

Grams of Fe= 1.6*56= 89.6 gms

Part D

Percent Yields = 100* actual yield/ theoretical yields = 100*63.2/89.6=70.53%

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.