Question

1. According to the following reaction, how many grams of barium chloride will be formed upon the complete reaction of 24.2 grams of barium hydroxide with excess hydrochloric acid?

hydrochloric acid (aq) + barium hydroxide (aq)--->barium chloride (aq) + water (l)

___grams barium chloride

2-

For the following reaction, 29.8 grams of diphosphorus pentoxide are allowed to react with 9.54 grams of water.

diphosphorus pentoxide (s) + water (l) ---> phosphoric acid (aq)

What is the maximum amount of phosphoric acid that can be formed? ___ grams

What is the FORMULA for the limiting reagent?

3-

For the following reaction, 63.9 grams of barium hydroxide are allowed to react with 40.5 grams of sulfuric acid.

barium hydroxide (aq) + sulfuric acid (aq) --->barium sulfate (s) + water (l)

What is the maximum amount of barium sulfate 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

What amount of the excess reagent remains after the reaction is complete?___ grams

Answer 1

1)

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 of Ba(OH)2 = 24.2 g

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

= 24.2/1.713*10^2

= 0.1413 mol

Balanced chemical equation is:

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

According to balanced equation

mol of BaCl2 formed = moles of Ba(OH)2

= 0.1413 mol

Molar mass of BaCl2,

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

= 1*137.3 + 2*35.45

= 208.2 g/mol

mass of BaCl2 = number of mol * molar mass

= 0.1413*2.082*10^2

= 29.41 g

Answer: 29.4 g

2)

a)

Molar mass of P2O5,

MM = 2*MM(P) + 5*MM(O)

= 2*30.97 + 5*16.0

= 141.94 g/mol

mass(P2O5)= 29.8 g

use:

number of mol of P2O5,

n = mass of P2O5/molar mass of P2O5

=(29.8 g)/(1.419*10^2 g/mol)

= 0.2099 mol

Molar mass of H2O,

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

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass(H2O)= 9.54 g

use:

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(9.54 g)/(18.02 g/mol)

= 0.5295 mol

Balanced chemical equation is:

P2O5 + 3 H2O ---> 2 H3PO4 +

1 mol of P2O5 reacts with 3 mol of H2O

for 0.2099 mol of P2O5, 0.6298 mol of H2O is required

But we have 0.5295 mol of H2O

so, H2O is limiting reagent

we will use H2O in further calculation

Molar mass of H3PO4,

MM = 3*MM(H) + 1*MM(P) + 4*MM(O)

= 3*1.008 + 1*30.97 + 4*16.0

= 97.994 g/mol

According to balanced equation

mol of H3PO4 formed = (2/3)* moles of H2O

= (2/3)*0.5295

= 0.353 mol

use:

mass of H3PO4 = number of mol * molar mass

= 0.353*97.99

= 34.59 g

Answer: 34.6 g

b)

Answer: H2O

3)

a)

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)= 63.9 g

use:

number of mol of Ba(OH)2,

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

=(63.9 g)/(1.713*10^2 g/mol)

= 0.373 mol

Molar mass of H2SO4,

MM = 2*MM(H) + 1*MM(S) + 4*MM(O)

= 2*1.008 + 1*32.07 + 4*16.0

= 98.086 g/mol

mass(H2SO4)= 40.5 g

use:

number of mol of H2SO4,

n = mass of H2SO4/molar mass of H2SO4

=(40.5 g)/(98.09 g/mol)

= 0.4129 mol

Balanced chemical equation is:

Ba(OH)2 + H2SO4 ---> BaSO4 + 2 H2O

1 mol of Ba(OH)2 reacts with 1 mol of H2SO4

for 0.373 mol of Ba(OH)2, 0.373 mol of H2SO4 is required

But we have 0.4129 mol of H2SO4

so, Ba(OH)2 is limiting reagent

we will use Ba(OH)2 in further calculation

Molar mass of BaSO4,

MM = 1*MM(Ba) + 1*MM(S) + 4*MM(O)

= 1*137.3 + 1*32.07 + 4*16.0

= 233.37 g/mol

According to balanced equation

mol of BaSO4 formed = (1/1)* moles of Ba(OH)2

= (1/1)*0.373

= 0.373 mol

use:

mass of BaSO4 = number of mol * molar mass

= 0.373*2.334*10^2

= 87.05 g

Answer: 87.1 g

b)

Answer: Ba(OH)2

1)

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 of Ba(OH)2 = 24.2 g

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

= 24.2/1.713*10^2

= 0.1413 mol

Balanced chemical equation is:

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

According to balanced equation

mol of BaCl2 formed = moles of Ba(OH)2

= 0.1413 mol

Molar mass of BaCl2,

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

= 1*137.3 + 2*35.45

= 208.2 g/mol

mass of BaCl2 = number of mol * molar mass

= 0.1413*2.082*10^2

= 29.41 g

Answer: 29.4 g

2)

a)

Molar mass of P2O5,

MM = 2*MM(P) + 5*MM(O)

= 2*30.97 + 5*16.0

= 141.94 g/mol

mass(P2O5)= 29.8 g

use:

number of mol of P2O5,

n = mass of P2O5/molar mass of P2O5

=(29.8 g)/(1.419*10^2 g/mol)

= 0.2099 mol

Molar mass of H2O,

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

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass(H2O)= 9.54 g

use:

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(9.54 g)/(18.02 g/mol)

= 0.5295 mol

Balanced chemical equation is:

P2O5 + 3 H2O ---> 2 H3PO4 +

1 mol of P2O5 reacts with 3 mol of H2O

for 0.2099 mol of P2O5, 0.6298 mol of H2O is required

But we have 0.5295 mol of H2O

so, H2O is limiting reagent

we will use H2O in further calculation

Molar mass of H3PO4,

MM = 3*MM(H) + 1*MM(P) + 4*MM(O)

= 3*1.008 + 1*30.97 + 4*16.0

= 97.994 g/mol

According to balanced equation

mol of H3PO4 formed = (2/3)* moles of H2O

= (2/3)*0.5295

= 0.353 mol

use:

mass of H3PO4 = number of mol * molar mass

= 0.353*97.99

= 34.59 g

Answer: 34.6 g

b)

Answer: H2O

3)

a)

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)= 63.9 g

use:

number of mol of Ba(OH)2,

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

=(63.9 g)/(1.713*10^2 g/mol)

= 0.373 mol

Molar mass of H2SO4,

MM = 2*MM(H) + 1*MM(S) + 4*MM(O)

= 2*1.008 + 1*32.07 + 4*16.0

= 98.086 g/mol

mass(H2SO4)= 40.5 g

use:

number of mol of H2SO4,

n = mass of H2SO4/molar mass of H2SO4

=(40.5 g)/(98.09 g/mol)

= 0.4129 mol

Balanced chemical equation is:

Ba(OH)2 + H2SO4 ---> BaSO4 + 2 H2O

1 mol of Ba(OH)2 reacts with 1 mol of H2SO4

for 0.373 mol of Ba(OH)2, 0.373 mol of H2SO4 is required

But we have 0.4129 mol of H2SO4

so, Ba(OH)2 is limiting reagent

we will use Ba(OH)2 in further calculation

Molar mass of BaSO4,

MM = 1*MM(Ba) + 1*MM(S) + 4*MM(O)

= 1*137.3 + 1*32.07 + 4*16.0

= 233.37 g/mol

According to balanced equation

mol of BaSO4 formed = (1/1)* moles of Ba(OH)2

= (1/1)*0.373

= 0.373 mol

use:

mass of BaSO4 = number of mol * molar mass

= 0.373*2.334*10^2

= 87.05 g

Answer: 87.1 g

b)

Answer: Ba(OH)2