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"#.
