Question

References Use the References to access important values if needed for this question. For the following reaction, 32.0 grams of calcium hydroxide are allowed to react with 27.9 grams of hydrochloric acid. calcium hydroxide (aq) + hydrochloric acid (aq) — calcium chloride (aq) + water (1) What is the maximum amount of calcium chloride that can be formed? grams What is the FORMULA for the limit eagent? What amount of the excess reagent remains after the reaction is complete? grams Submit Answer

Answer 1

1)

Molar mass of Ca(OH)2,

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

= 1*40.08 + 2*16.0 + 2*1.008

= 74.096 g/mol

mass(Ca(OH)2)= 32.0 g

use:

number of mol of Ca(OH)2,

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

=(32 g)/(74.1 g/mol)

= 0.4319 mol

Molar mass of HCl,

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

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass(HCl)= 27.9 g

use:

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(27.9 g)/(36.46 g/mol)

= 0.7653 mol

Balanced chemical equation is:

Ca(OH)2 + 2 HCl ---> CaCl2 + 2 H2O

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

for 0.4319 mol of Ca(OH)2, 0.8637 mol of HCl is required

But we have 0.7653 mol of HCl

so, HCl is limiting reagent

we will use HCl in further calculation

Molar mass of CaCl2,

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

= 1*40.08 + 2*35.45

= 110.98 g/mol

According to balanced equation

mol of CaCl2 formed = (1/2)* moles of HCl

= (1/2)*0.7653

= 0.3826 mol

use:

mass of CaCl2 = number of mol * molar mass

= 0.3826*1.11*10^2

= 42.46 g

Answer: 42.5 g

2)

HCl is limiting reagent

Answer: HCl

3)

According to balanced equation

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

= (1/2)*0.7653

= 0.3826 mol

mol of Ca(OH)2 remaining = mol initially present - mol reacted

mol of Ca(OH)2 remaining = 0.4319 - 0.3826

mol of Ca(OH)2 remaining = 4.924*10^-2 mol

Molar mass of Ca(OH)2,

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

= 1*40.08 + 2*16.0 + 2*1.008

= 74.096 g/mol

use:

mass of Ca(OH)2,

m = number of mol * molar mass

= 4.924*10^-2 mol * 74.1 g/mol

= 3.648 g

Answer: 3.65 g