answersLogoWhite

0


Best Answer

150 M NaOH = 150 mole / liter

20.0 mL = 0.0200 L

150 * 0.0200 = 3 moles

User Avatar

Wiki User

13y ago
This answer is:
User Avatar

Add your answer:

Earn +20 pts
Q: If 20.0 mL of 150 M NaOH are used in a titration how many moles of NaOH are used?
Write your answer...
Submit
Still have questions?
magnify glass
imp
Continue Learning about Natural Sciences

If 5.0 moles of NH3 are produce how many moles of N2 must have been used?

If 5.0 moles of NH3 are produced 2.5 moles of N2 are used.


What happens when you forget to remove the air bubble from the tip of the buret that contains NaOH in a titration?

Your results are likely to be slightly off. How much off depends on how big the air bubble was and how much titrant you actually used.


What volume of 0.150 m hclo4 solution is needed to neutralize 60 ml 0f 8.6510-2m naoh?

You MUST get the chemical equation for this question, which is: HCL + NaOH --> H20 + NaCl This equation is already balanced. This is a double displacement reaction with neutralization meaning water and a salt will be the products. M = n/V (molarity = number of moles/ volume other re-arrangements of this formula: n = MV V = n/M Molarity is always moles per litre, so Its good to change all given volumes to litres then convert back to ml after if you have to. 42.0ml can be converted to litres: 42.0ml = 0.042L You now have VOLUME and MOLARITY of NaOH, now used n = MV to get number of moles: 0.042 x 0.150 = 0.0063mol NaOH now you know number of moles, and a BALANCED chemical equation, so you can now do a stoicheometry conversion factor to get the number of moles for HCl. 0.0063mol NaOH x1mol HCL ÷ 1mol NaOH = 0.0063 mol (moles of NaOH cancel out and now you have # of moles for HCl. at this point you know HCL has Molarity 0.126 and 0.0063mol. now solve for volume: V = n/M V = 0.0063/0.126 V = 0.05L You now know Volume of HCL is 0.05L, now convert it mL because the question is asking for it. 0.05L = 0.05 x 1000 = 50ml Question complete! *note* in this chemical equation, the number of moles is the same because the chemical equation is a 1:1 ratio, meaning their are no coefficients in front of an substance. is the HCl or NaOH had coefficients, then the answer would be completly different, thats why you need the stoicheometry mole to mole conversion factor that I added in.


How many moles potassium chlorate must be used to produce 6 moles of oxygen gas?

We need 3 moles of potassium perchlorate.


How many moles of potassium chlorate must be used to produce 6 moles if oxygen gas?

Four moles of potassium chlorate are needed.

Related questions

What volume of NaOH was used in the titration?

Why all peoples are selfish


Which indicator is used in titration of naoh and h2so4?

Methyl orange


Why only potassium biphthalate is used for standerdization of NaOH?

acid-base titration


How do you test hydrochloric acid whether it is strong or weak?

What you are doing here is titration. You know you have a solution of HCl, but you do not know how much HCl is in it. For this you use something that can react with HCl (NaOH) and use an indicator to tell you when the reaction is complete. The reaction is pretty simple: HCl + NaOH --> H2O + NaCl You can see here that NaOH and HCl have a 1:1 mol relationship. So, lets find out how many moles of NaOH you used up with concentration = moles/volume 0.10 M NaOH = moles NaOH/ 0.0197 L NaOH solution Remember that M is in moles/L moles NaOH = 0.00197 moles Since you have a 1:1 relationship of NaOH with HCl, the 0.00197 mol applies to HCl as well. The next question works the same way, but backwards. Try doing it yourself if you understood the first part before reading my answer. Find out how many moles of HCl you have so you can find out how much moles of NaOH you need for the neutralization. 0.050 M HCl = n HCl / 0.020 L HCl soln n HCl = 0.001 mol HCl Remeber the 1:1 relationship, which gives you that n NaOH = 0.001 mol. Now all you need is the volume. 0.1 M NaOH = 0.001 mol NaOH/ Volume soln V = 0.01 L = 10 mL


Why Gycerol is used for the titration of NaOH with Borax?

cuz' borax contain boric acid (H3BO3) which is very weak acid , thus it gives non accurate titration, so we add neutral glycerol to increase the acidity by the way we add glycerol after titration of NaOH not before


In HCl and NaOH titration which indicator is used?

Phenolphthalein indicates red for acids and blue for bases.


What unknown quantity can be calculated after performing a titration?

Titrations can be used to work out the initial amount of moles of a substance (for instance the number of moles of iron in a tablet).


Which indicator is used for titration of C2H2O4 with NaOH?

Phenolphtalein change colorless at pH < 8 to purple blue at pH > 8 to 10


What chemical test is used to measure citric acid content in drinks?

Some methods are: absorption spectrophotometry, amperometry, titration with NaOH.


Why is thymol blue used as the indicator for titration of acetic acid and naoh?

An indicator that changes its colour around pH value 5 should be used in this titration. Alternatively, instead of Thymol Blue, Methyl Orange or Methyl Red may be used.


How many moles of sodium hydroxide are formed when 46 g of sodium react with water?

First construct a a balanced equation for the reaction : 2Na + 2H2O --> 2NaOH + H2 From the equation, we know that the ratio of Sodium(Na) to the ratio of Sodium Hydroxide(NaOH) is the same. Ar of Na = 23g/mol Number of moles = mass / Ar = 46g / 23g/mol = 2mol (moles of Na used) Since ratio of Na to NaOH is the same, therefore there are also 2 moles of NaOH formed.


What is the concentration of excess H2SO4when50cm3 of 0.200moldm-3 NaOH is added to 40cm3 of 0.250 moldm-3 H2SO4.?

moles H2SO4 = 0.040 L x 0.250 mol/L = 0.01 moles H2SO4moles NaOH = 0.05 L x 0.200 mol/L = 0.01 moles NaOHReaction: 2NaOH + H2SO4 = Na2SO4 + 2H2OIt takes 2 moles NaOH for each 1 mole H2SO4Moles H2SO4 used up by 0.01 moles NaOH = 0.005 molesmoles H2SO4 remaining = 0.005 moles H2SO4Total volume = 50 cm3 + 40 cm3 = 90 cm3 = 0.090 LFinal [H2SO4] = 0.005 moles/0.090 L = 0.055 M (to 2 significant figures)