Respuesta :
Answer: A. 3.28 g of HCl
B. 6.30 g of [tex]HNO_3[/tex]
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} NaOH=\frac{3.65g}{40g/mol}=0.09moles[/tex]
[tex]\text{Moles of} Ca(OH)_2=\frac{3.65g}{74g/mol}=0.05moles[/tex]
a) [tex]HCl(aq)+NaOH(aq)\rightarrow H_2O(l)+NaCl(aq)[/tex]
According to stoichiometry :
1 mole of [tex]NaOH[/tex] require = 1 mole of [tex]HCl[/tex]
Thus 0.09 moles of [tex]NaOH[/tex] will require=[tex]\frac{1}{1}\times 0.09=0.09moles[/tex] of [tex]HCl[/tex]
Mass of [tex]HCl=moles\times {\text {Molar mass}}=0.09moles\times 36.5g/mol=3.28g[/tex]
b) [tex]2HNO_3(aq)+Ca(OH)_2(aq)\rightarrow 2H_2O(l)+Ca(NO_3)_2(aq)[/tex]
According to stoichiometry :
1 mole of [tex]Ca(OH)_2[/tex] require = 2 moles of [tex]HNO_3[/tex]
Thus 0.05 moles of [tex]Ca(OH)_2[/tex] will require=[tex]\frac{2}{1}\times 0.05=0.1moles[/tex] of [tex]HNO_3[/tex]
Mass of [tex]HNO_3=moles\times {\text {Molar mass}}=0.1moles\times 63g/mol=6.3g[/tex]
