Question

For each of the following acid-base reactions,

calculate the mass (in grams) of each acid necessary
to completely react with and neutralize 3.65 g of the
base.
Part A
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)

Part B
2 HNO3(aq) + Ca(OH)2(aq)
2 H2O(1) + C

Answer 1

Answer: A. 3.28 g of HCl

B. 6.30 g of
`HNO_3`

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} NaOH=(3.65g)/(40g/mol)=0.09moles`

`\text{Moles of} Ca(OH)_2=(3.65g)/(74g/mol)=0.05moles`

a)
`HCl(aq)+NaOH(aq)\rightarrow H_2O(l)+NaCl(aq)`

According to stoichiometry :

1 mole of
`NaOH` require = 1 mole of
`HCl`

Thus 0.09 moles of
`NaOH` will require=
`(1)/(1)* 0.09=0.09moles` of
`HCl`

Mass of
`HCl=moles* {\text {Molar mass}}=0.09moles* 36.5g/mol=3.28g`

b)
`2HNO_3(aq)+Ca(OH)_2(aq)\rightarrow 2H_2O(l)+Ca(NO_3)_2(aq)`

According to stoichiometry :

1 mole of
`Ca(OH)_2` require = 2 moles of
`HNO_3`

Thus 0.05 moles of
`Ca(OH)_2` will require=
`(2)/(1)* 0.05=0.1moles` of
`HNO_3`

Mass of
`HNO_3=moles* {\text {Molar mass}}=0.1moles* 63g/mol=6.3g`