In: Chemistry
For each of the following acid-base reactions, calculate how many grams of each acid are necessary to completely react with and neutralize 2.7 g of the base.
A. HCl(aq)+NaOH(aq)→H2O(l)+NaCl(aq)
B. 2HNO3(aq)+Ca(OH)2(aq)→2H2O(l)+Ca(NO3)2(aq)
C. H2SO4(aq)+2KOH(aq)→2H2O(l)+K2SO4(aq)
ans)
from aboce data that
Part A
HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)
molar mass of NaOH = 39.997 g/mol
no. of mole = mass of substance in gm / molar mass
no. of mole of NaOH = 2.7/ 39.997 = 0.0675 mole
According to reaction to 1 mole of NaOH required HCl is 1 mole they react in 1:1 proportion therefore to neutrilise 0.0675 mole of Base NaOH required acid HCl = 0.0675 mole
HCl is required = 0.0675 mole
molar mass of HCl = 36.46 g/mol
mass of compound in gm = no. of mole X molar mass
mass of acid HCl requied = 0.0675 X 36.46 = 2.46 gm
mass of acid HCl requied = 2.46 gm
Part B
2HNO3(aq) + Ca(OH)2(aq) 2H2O(l) + Ca(NO3)2(aq)
molar mass of Ca(OH)2 = 74.093 g/mol
no. of mole = mass of substance in gm / molar mass
no. of mole of Ca(OH)2 = 2.7/ 74.093 = 0.03644 mole
According to reaction to 1 mole of Ca(OH)2 required HNO3 is 2 mole they react in 1:2 proportion therefore to neutrilise 0.03644 mole of Base Ca(OH)2 required acid HNO3 = 0.03644X 2 = 0.07288mole
HNO3 is required = 0.07288 mole
molar mass of HNO3 = 63.01 g/mol
mass of compound in gm = no. of mole X molar mass
mass of acid HNO3 requied = 0.07288 X 63.01 = 4.59 gm
mass of acid HNO3 requied = 4.59 gm
Part C
H2SO4(aq) + 2KOH(aq) 2H2O(l) + K2SO4(aq)
molar mass of KOH = 56.1056 g/mol
no. of mole = mass of substance in gm / molar mass
no. of mole of KOH =2.7 / 56.1056 = 0.048 mole
According to reaction to 2 mole of KOH required H2SO4 is 1 mole they react in 2:1 proportion therefore to neutrilise 0.09 mole of Base KOH required acid H2SO4 = 0.048/2 = 0.024 mole
H2SO4 is required = 0.024 mole
molar mass of H2SO4 = 98.079 g/mol
mass of compound in gm = no. of mole X molar mass
mass of acid H2SO4 requied = 0.024 X 98.079 =2.35 gm
mass of acid H2SO4 requied = 2.35 gm