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
YOU DON'T NEED TO ANSWER ALL THE QUESTIONS ALL I NEED HELP WITH IS QUESTIONS 4 and 5. I got d, e, e for the 1st 3 questions if that helps you, but you don't need to worry about questions 1-3 I just included them in case it helps you.
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
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