Question

5.00 g of solid sodium and 30.0 g of liquid bromine react to form solid NaBr.

Sodium is the limiting reactant.

2Na + Br₂ ---> 2NaBr

a. How many grams of excess reactant are left over at the end of the reaction?

Show all work.

Answer 1

Molar mass of Na = 22.99 g/mol

mass of Na = 5.0 g

we have below equation to be used:

number of mol of Na,

n = mass of Na/molar mass of Na

=(5.0 g)/(22.99 g/mol)

= 0.2175 mol

Molar mass of Br2 = 159.8 g/mol

mass of Br2 = 30.0 g

we have below equation to be used:

number of mol of Br2,

n = mass of Br2/molar mass of Br2

=(30.0 g)/(159.8 g/mol)

= 0.1877 mol

we have the Balanced chemical equation as:

2 Na + Br2 ---> 2 NaBr +

2 mol of Na reacts with 1 mol of Br2

for 0.2175 mol of Na, 0.1087 mol of Br2 is required

But we have 0.1877 mol of Br2

so, Na is limiting reagent

we will use Na in further calculation

From balanced chemical reaction, we see that

when 2 mol of Na reacts, 1 mol of Br2 is formed

mol of Br2 reacted = (1/2)* moles of Na

= (1/2)*0.2175

= 0.1087 mol

mol of Br2 remaining = mol initially present - mol reacted

mol of Br2 remaining = 0.1877 - 0.1087

mol of Br2 remaining = 0.079 mol

Molar mass of Br2 = 159.8 g/mol

we have below equation to be used:

mass of Br2,

m = number of mol * molar mass

= 7.899*10^-2 mol * 159.8 g/mol

= 12.6 g

Answer: 12.6 g