k,S04 (aq) +2H20 (1) K 34,102 32, oc O lex4 5) Solid potassium hydroxide reactin
ID: 717160 • Letter: K
Question
k,S04 (aq) +2H20 (1) K 34,102 32, oc O lex4 5) Solid potassium hydroxide reacting with sulfuric acid to produce aqueous potassium 1 74. sulfate and liquid water. a. If 25.g of sulfuric acid is reacted with 7.7 g of potassium hydroxide, how many grams of potassium sulfate are produced?SC of excess reactant that remains after the reaction is completed. would you expect to obtain if the reaction has a 47.2% yield? b. For part (a) of this problem, identify the limiting reactant and calculate the mass c. From your theoretical yield calculated in part (a), how many grams of materialExplanation / Answer
a)
Molar mass of KOH,
MM = 1*MM(K) + 1*MM(O) + 1*MM(H)
= 1*39.1 + 1*16.0 + 1*1.008
= 56.108 g/mol
mass(KOH)= 7.7 g
use:
number of mol of KOH,
n = mass of KOH/molar mass of KOH
=(7.7 g)/(56.11 g/mol)
= 0.1372 mol
Molar mass of H2SO4,
MM = 2*MM(H) + 1*MM(S) + 4*MM(O)
= 2*1.008 + 1*32.07 + 4*16.0
= 98.086 g/mol
mass(H2SO4)= 25.0 g
use:
number of mol of H2SO4,
n = mass of H2SO4/molar mass of H2SO4
=(25.0 g)/(98.09 g/mol)
= 0.2549 mol
Balanced chemical equation is:
2 KOH + H2SO4 ---> K2SO4 + 2 H2O
2 mol of KOH reacts with 1 mol of H2SO4
for 0.1372 mol of KOH, 6.862*10^-2 mol of H2SO4 is required
But we have 0.2549 mol of H2SO4
so, KOH is limiting reagent
we will use KOH in further calculation
Molar mass of K2SO4,
MM = 2*MM(K) + 1*MM(S) + 4*MM(O)
= 2*39.1 + 1*32.07 + 4*16.0
= 174.27 g/mol
According to balanced equation
mol of K2SO4 formed = (1/2)* moles of KOH
= (1/2)*0.1372
= 6.862*10^-2 mol
use:
mass of K2SO4 = number of mol * molar mass
= 6.862*10^-2*1.743*10^2
= 11.96 g
Answer: 12.0 g
b)
KOH is limiting reagent
According to balanced equation
mol of H2SO4 reacted = (1/2)* moles of KOH
= (1/2)*0.1372
= 6.862*10^-2 mol
mol of H2SO4 remaining = mol initially present - mol reacted
mol of H2SO4 remaining = 0.2549 - 6.862*10^-2
mol of H2SO4 remaining = 0.1863 mol
Molar mass of H2SO4,
MM = 2*MM(H) + 1*MM(S) + 4*MM(O)
= 2*1.008 + 1*32.07 + 4*16.0
= 98.086 g/mol
use:
mass of H2SO4,
m = number of mol * molar mass
= 0.1863 mol * 98.09 g/mol
= 18.27 g
Answer: 18.3 g
c)
% yield = actual mass*100/theoretical mass
47.2= actual mass*100/11.96
actual mass=5.644 g
Answer: 5.64 g
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