Question

For the following reaction, 29.4 grams of hydrochloric acid are allowed to react with 59.3 grams of barium hydroxide. hydrochloric acid (aq) + barium hydroxide (aq) — barium chloride (aq) + water (1) What is the maximum amount of barium chloride that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

1)

a)

Molar mass of HCl,

MM = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass(HCl)= 29.4 g

use:

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(29.4 g)/(36.46 g/mol)

= 0.8064 mol

Molar mass of Ba(OH)2,

MM = 1*MM(Ba) + 2*MM(O) + 2*MM(H)

= 1*137.3 + 2*16.0 + 2*1.008

= 171.316 g/mol

mass(Ba(OH)2)= 59.2 g

use:

number of mol of Ba(OH)2,

n = mass of Ba(OH)2/molar mass of Ba(OH)2

=(59.2 g)/(1.713*10^2 g/mol)

= 0.3456 mol

Balanced chemical equation is:

2 HCl + Ba(OH)2 ---> BaCl2 + 2 H2O

2 mol of HCl reacts with 1 mol of Ba(OH)2

for 0.8064 mol of HCl, 0.4032 mol of Ba(OH)2 is required

But we have 0.3456 mol of Ba(OH)2

so, Ba(OH)2 is limiting reagent

we will use Ba(OH)2 in further calculation

Molar mass of BaCl2,

MM = 1*MM(Ba) + 2*MM(Cl)

= 1*137.3 + 2*35.45

= 208.2 g/mol

According to balanced equation

mol of BaCl2 formed = (1/1)* moles of Ba(OH)2

= (1/1)*0.3456

= 0.3456 mol

use:

mass of BaCl2 = number of mol * molar mass

= 0.3456*2.082*10^2

= 71.95 g

Answer: 72.0 g

b)

Ba(OH)2 is limiting reagent

Answer: Ba(OH)2

c)

According to balanced equation

mol of HCl reacted = (2/1)* moles of Ba(OH)2

= (2/1)*0.3456

= 0.6911 mol

mol of HCl remaining = mol initially present - mol reacted

mol of HCl remaining = 0.8064 - 0.6911

mol of HCl remaining = 0.1153 mol

Molar mass of HCl,

MM = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.45

= 36.458 g/mol

use:

mass of HCl,

m = number of mol * molar mass

= 0.1153 mol * 36.46 g/mol

= 4.203 g

Answer: 4.20 g

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