Question

Question 19 of 24 > A sample of 6.11 g of solid calcium hydroxide is added to 32.0 mL of 0.340 M aqueous hydrochloric acid. Write the balanced chemical equation for the reaction. Physical states are optional. chemical equation: What is the limiting reactant? calcium hydroxide hydrochloric acid How many grams of salt are formed after the reaction is complete? mass of salt:

Answer 1

1)

Balanced chemical equation is:

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

2)

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)= 6.11 g

use:

number of mol of Ca(OH)2,

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

=(6.11 g)/(74.1 g/mol)

= 8.246*10^-2 mol

volume of HCl, V = 32 mL

= 3.2*10^-2 L

use:

number of mol in HCl,

n = Molarity * Volume

= 0.34*3.2*10^-2

= 1.088*10^-2 mol

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

for 8.246*10^-2 mol of Ca(OH)2, 0.1649 mol of HCl is required

But we have 1.088*10^-2 mol of HCl

so, HCl is limiting reagent

Answer: hydrochloric acid.

3)

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)*1.088*10^-2

= 5.44*10^-3 mol

use:

mass of CaCl2 = number of mol * molar mass

= 5.44*10^-3*1.11*10^2

= 0.6037 g

Answer: 0.604 g