n=c/v
n=3M/.25L
n=12 mol
m=Mxn
m=58.443 g/mol x 12 mol
m=701.3 g
n= mol
c=concentration
v=volume
m=mass
M= molar mass
Tylerops:
I don't agree with this answer. Molarity is defined as Moles/Liters. In other words Molarity is the concentration of a solution. In the above n= Concentration / Liters. That is equal to saying Moles=(Moles/liters)/ Liters. In the above question the concentration is (3 moles/ liter), or 3M.
Plus, how can it be possible to have 12 moles in 250ml when you only have 3 moles in each liter of the original solution?
Correct ANSWER:
3.00 M, or 3 moles per (L) "liter" calls for having 3 moles per liter of the solution. The question asks how many moles must be in 250ml of a solution that has 3 moles per Liter.
You must ask yourself what percent of 1 Liter is 250mls? Since there are a thousand ml in one liter, (1000ml=1L), then 250ml is exactly 25% of a Liter, or .25L.
So, 250ml can only hold 25% of the 3.00 Molarity. Meaning that you multiply 3 x .25 and get .75 moles. 58.443g/molNaCl x .75 moles = FINAL ANSWER 43.83225g NaCl, Sig Fig, 43.83gNaCl
0.1 M sodium hydroxide solution contains 0.1 moles of sodium hydroxide per liter of solution. This corresponds to 0.1 moles of NaOH per 40 g (1 mole) of NaOH, resulting in 4 g of NaOH in 1 liter of 0.1 M NaOH solution.
The molar mass of sodium hydroxide (NaOH) is approximately 40 grams per mole. Therefore, a 6.94 mole sample of sodium hydroxide would contain approximately 278 grams (6.94 moles x 40 grams/mole).
Na +H2O -> NaOH +(1/2)H2 Every mole of Sodium requires one mole of water to make one mole of Sodium Hydroxide. So two moles of Sodium will produce two moles of Sodium Hydroxide. If there are three moles of water in the initial reaction then there will be one mole of water left over after reacting with two moles of Sodium. This reaction will produce half a mole of hydrogen gas.
To calculate the grams of sodium hydroxide present in the solution, first calculate the number of moles using the formula: moles = Molarity (M) x Volume (L). Then, use the molar mass of sodium hydroxide (NaOH) to convert moles to grams. The molar mass of NaOH is 40 g/mol. Thus, in this case, you have 0.3375 moles of NaOH and if you convert this to grams, it would be 13.5 grams.
In the acid-base reaction where sodium hydroxide and sulfuric acid react, the formula is: H2SO4 + 2NaOH --> Na2SO4 + 2H2O. The coefficients shown are necessary to uphold the law of conservation of mass. So, if you have 17 moles of sulfuric acid, you will need twice as many moles of sodium hydroxide, so the answer is 34 moles NaOH.
A 3 M sodium hydroxide solution means there are 3 moles of sodium hydroxide dissolved in 1 liter of solution.
Molarity means moles per litre. So here you have to divide the moles by 75 and multiply by 1000. 0.5/75*1000, which is 6.667 molar.
0.1 M sodium hydroxide solution contains 0.1 moles of sodium hydroxide per liter of solution. This corresponds to 0.1 moles of NaOH per 40 g (1 mole) of NaOH, resulting in 4 g of NaOH in 1 liter of 0.1 M NaOH solution.
First, calculate the number of moles of nitric acid present in 3.50 L of 0.700 M solution. Since nitric acid is a diprotic acid, the mole ratio with sodium hydroxide is 1:2. Then, use the mole ratio to determine the number of moles of sodium hydroxide needed to neutralize the nitric acid. Finally, convert the moles of sodium hydroxide to grams using its molar mass.
Sodium in its elemental form is just sodium metal, Na. Thus assuming that the sample of sodium is pure, there are 2.50 moles of sodium in a 2.50mol sample.
The molar mass of sodium hydroxide (NaOH) is approximately 40 grams per mole. Therefore, a 6.94 mole sample of sodium hydroxide would contain approximately 278 grams (6.94 moles x 40 grams/mole).
Na +H2O -> NaOH +(1/2)H2 Every mole of Sodium requires one mole of water to make one mole of Sodium Hydroxide. So two moles of Sodium will produce two moles of Sodium Hydroxide. If there are three moles of water in the initial reaction then there will be one mole of water left over after reacting with two moles of Sodium. This reaction will produce half a mole of hydrogen gas.
To calculate the grams of sodium hydroxide present in the solution, first calculate the number of moles using the formula: moles = Molarity (M) x Volume (L). Then, use the molar mass of sodium hydroxide (NaOH) to convert moles to grams. The molar mass of NaOH is 40 g/mol. Thus, in this case, you have 0.3375 moles of NaOH and if you convert this to grams, it would be 13.5 grams.
In the acid-base reaction where sodium hydroxide and sulfuric acid react, the formula is: H2SO4 + 2NaOH --> Na2SO4 + 2H2O. The coefficients shown are necessary to uphold the law of conservation of mass. So, if you have 17 moles of sulfuric acid, you will need twice as many moles of sodium hydroxide, so the answer is 34 moles NaOH.
Molarity = moles of solute/Liters of solution ( 1500 mL = 1.5 Liters ) Molarity = 0.800 moles NaOH/1.5 Liters = 0.533 M sodium hydroxide ...
No amount of sodium sulphate can be formed from sodium hydroxide alone, because sodium sulfate contains sulfur and sodium hydroxide does not. By neutralization with sulphuric acid, one formula unit of sodium sulphate can be formed from two moles of sodium hydroxide, according to the equation 2 NaOH + H2SO4 -> Na2SO4 + 2 H2O.
You can calculate the concentration of a phosphoric acid solution by determining the volume of sodium hydroxide needed to neutralize it in a titration. The molarity of the sodium hydroxide solution and the balanced chemical equation for the reaction will allow you to find the moles of phosphoric acid present, hence the concentration.