1.What reagent is used as the indicator for the neutralization? a)citric acid b)sodium hydroxide c)sodium citrate d)phenolphthalein 2. What is the mole ratio of NaOH to citric acid for this neutralization reaction? a)2 mole NaOH/1 mole citric acid b)1 mole NaOH/1 mole citric acid c)1 mole NaOH/2 mole citric acid d)1 mole NaOH/3 mole citric acid e)3 mole NaOH/1 mole citric acid 3.If 35.00 mL of 0.1104 M NaOH were used to reach the neutralization point, how many moles of NaOH were dispensed? a)3.154 x 10-3 mol NaOH b)1.104 x 10-4 mol NaOH c)258.8 mol NaOH d)3.864 x 10-3 mol NaOH
1.What reagent is used as the indicator for the neutralization?
a)citric acid
b)sodium hydroxide
c)sodium citrate
d)phenolphthalein
2.
What is the mole ratio of NaOH to citric acid for this neutralization reaction?
a)2 mole NaOH/1 mole citric acid
b)1 mole NaOH/1 mole citric acid
c)1 mole NaOH/2 mole citric acid
d)1 mole NaOH/3 mole citric acid
e)3 mole NaOH/1 mole citric acid
3.If 35.00 mL of 0.1104 M NaOH were used to reach the neutralization point, how many moles of NaOH were dispensed?
a)3.154 x 10-3 mol NaOH
b)1.104 x 10-4 mol NaOH
c)258.8 mol NaOH
d)3.864 x 10-3 mol NaOH
4. Using the above mole ratio, convert the moles of NaOH in the previous question to moles of citric acid.
a)1.288 x 10-3 mol citric acid
b)1.159 x 10-2 mol citric acid
c)3.864 x 10-3 mol citric acid
d)1.932 x 10-3 mol citric acid
5.
How many grams of citric acid were neutralized in the question above? (MM = 192.12 g/mol)
a)0.7425 g citric acid
b)0.2475 g citric acid
c)0.0825 g citric acid
d)6.704 x 10-6 g citric acid
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 3 images