Question #6b9a2

1 Answer
May 25, 2017

Answer:

#"a."# #0.02# #"mol"#

#"b."# #0.89# #"g"#

Explanation:

In order to answer these questions we first need to know the equation representing this chemical reaction.

In this case, it is a reaction between #"HCl"# and #"Ca(OH)"_(2) :#

#"Ca(OH)"#2 #+# #"HCl"# #rightarrow# #"CaCl"#2 #+##"H"#2#"O"#

Let's balance the equation:

#"Ca(OH)"#2 #+# 2#"HCl"# #rightarrow# #"CaCl"#2 #+# 2#"H"#2#"O"#

#"a."# Using the formula #n = frac(m)(M):#

#Rightarrow n("Ca(OH)") = frac(0.75 " g")(74.10 " g mol"^(- 1))#

#Rightarrow n("Ca(OH)") = 0.01# #"mol"#

In the balanced chemical equation above, the stoichiometric ratio between #"HCl"# and #"Ca(OH)"# is #2 : 1#.

#therefore n("HCl") = 2 times 0.01# #"mol"#

#Rightarrow n("HCl") = 0.02# #"mol"#

#"b."# Let's set up a ratio using the molecular masses of #"Ca(OH)"# and #"CaCl"#:

#Rightarrow frac("MM of Ca(OH)")("MM of CaCl") = frac("m of Ca(OH)")("m of CaCl")#

#Rightarrow frac(74.10 " g mol"^(- 1))(110.98 "g mol"^(- 1)) = frac(0.75 " g")(x)#

We need to find #x#, which is the mass of #"CaCl"# formed in the reaction:

#Rightarrow 0.6676878717 = frac(0.75 " g")(x)#

#therefore x = 0.89# #"g"#

Therefore, the number of moles of #"HCl"# needed to completely react with #"Ca(OH)" "_(2)# is #0.02# #"mol"#, while the mass of #"CaCl"# formed in the reaction is #0.89# #"g"#.