Question

100.0-mL of 3.0M hydrobromic acid reacts with 75.0-mL of 3.0M sodium hydroxide to produce a salt and water. Write the balanced equation for this reaction and determine the max amount, in grams, of the salt that can form.

Answer 1

volume of HBr, V = 100.0 mL

= 0.1 L

we have below equation to be used:

number of mol in HBr,

n = Molarity * Volume

= 3*0.1

= 0.3 mol

volume of NaOH, V = 75.0 mL

= 7.5*10^-2 L

we have below equation to be used:

number of mol in NaOH,

n = Molarity * Volume

= 3*0.075

= 0.225 mol

we have the Balanced chemical equation as:

HBr + NaOH ---> NaBr + H2O

1 mol of HBr reacts with 1 mol of NaOH

for 0.3 mol of HBr, 0.3 mol of NaOH is required

But we have 0.225 mol of NaOH

so, NaOH is limiting reagent

we will use NaOH in further calculation

Molar mass of NaBr = 1*MM(Na) + 1*MM(Br)

= 1*22.99 + 1*79.9

= 102.89 g/mol

From balanced chemical reaction, we see that

when 1 mol of NaOH reacts, 1 mol of NaBr is formed

mol of NaBr formed = (1/1)* moles of NaOH

= (1/1)*0.225

= 0.225 mol

we have below equation to be used:

mass of NaBr = number of mol * molar mass

= 0.225*1.029*10^2

= 23.15 g

Answer: 23.15 g