PART A: A volume of 80.0mL of aqueous potassium hydroxide (KOH) was titrated aga
ID: 835004 • Letter: P
Question
PART A:
A volume of 80.0mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 25.7mL of 1.50 M H2SO4 was needed? The equation is 2KOH(aq)+H2SO4(aq)?K2SO4(aq)+2H2O(l)
molarity = .........??
PART B:
A volume of 80.0mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 25.7mL of 1.50 M H2SO4 was needed? The equation is 2KOH(aq)+H2SO4(aq)?K2SO4(aq)+2H2O(l)
mass of H2O2= .......... ??
can someone help me to solve this? :-/ :-/
Explanation / Answer
1) M1V1/n1=M2V2/n2
1.50*25.7/1=M2* 80/2
M2=0.963M
2) M1V1/n1=M2V2/n2
M1*80/2= 25.7*1.50/1
M2=0.963M
2 MOLES OF WATER IS PRODUCED
n=wt/m.wt
2=wt/18
wt=36 grams of water is produced
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