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 2.8 g of the base.
Part A : HCl(aq)+NaOH(aq)→H2O(l)+NaCl(aq)
Part B : 2HNO3(aq)+Ca(OH)2(aq)→2H2O(l)+Ca(NO3)2(aq)
Part C : H2SO4(aq)+2KOH(aq)→2H2O(l)+K2SO4(aq)
Express your answer using two significant figures.
A)
Molar mass of NaOH = 1*MM(Na) + 1*MM(O) + 1*MM(H)
= 1*22.99 + 1*16.0 + 1*1.008
= 39.998 g/mol
mass of NaOH = 2.8 g
mol of NaOH = (mass)/(molar mass)
= 2.8/39.998
= 0.0700 mol
From balanced chemical reaction, we see that
when 1 mol of NaOH reacts, 1 mol of HCl is formed
mol of HCl formed = moles of NaOH
= 0.07 mol
Molar mass of HCl = 1*MM(H) + 1*MM(Cl)
= 1*1.008 + 1*35.45
= 36.458 g/mol
mass of HCl = number of mol * molar mass
= 0.07*36.458
= 2.55 g
Answer: 2.6 g
B)
Molar mass of Ca(OH)2 = 1*MM(Ca) + 2*MM(O) + 2*MM(H)
= 1*40.08 + 2*16.0 + 2*1.008
= 74.096 g/mol
mass of Ca(OH)2 = 2.8 g
mol of Ca(OH)2 = (mass)/(molar mass)
= 2.8/74.096
= 0.0378 mol
From balanced chemical reaction, we see that
when 1 mol of Ca(OH)2 reacts, 2 mol of HNO3 is formed
mol of HNO3 formed = (2/1)* moles of Ca(OH)2
= (2/1)*0.0378
= 0.0756 mol
Molar mass of HNO3 = 1*MM(H) + 1*MM(N) + 3*MM(O)
= 1*1.008 + 1*14.01 + 3*16.0
= 63.018 g/mol
mass of HNO3 = number of mol * molar mass
= 0.0756*63.018
= 4.76 g
Answer: 4.8 g
C)
Molar mass of KOH = 1*MM(K) + 1*MM(O) + 1*MM(H)
= 1*39.1 + 1*16.0 + 1*1.008
= 56.108 g/mol
mass of KOH = 2.8 g
mol of KOH = (mass)/(molar mass)
= 2.8/56.108
= 0.0499 mol
From balanced chemical reaction, we see that
when 2 mol of KOH reacts, 1 mol of H2SO4 is formed
mol of H2SO4 formed = (1/2)* moles of KOH
= (1/2)*0.0499
= 0.025 mol
Molar mass of H2SO4 = 2*MM(H) + 1*MM(S) + 4*MM(O)
= 2*1.008 + 1*32.07 + 4*16.0
= 98.086 g/mol
mass of H2SO4 = number of mol * molar mass
= 0.025*98.086
= 2.45 g
Answer: 2.5 g