Q: How many moles of sodium hydroxide are in a 36.65g sample of NaOH?

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60 g NaOH x 1 mole NaOH/40 g NaOH = 1.5 moles NaOH

The number of moles is 1,8.

20 moles of NaOH needed to neutralize 20 moles of nitric acid

C2H4O2 + NaOH = H2O + C2H3O2Na Acetic acid (60 gm) + sodium hydroxide ( 40 gm) = 100 gm water (18 gm) + sodium acetate (82 gm) = 100 gm Ratio reactants to products = 1:1 Molarity = moles / L, 3M = 3 moles / 1 L Acetic acid = 60 gm / total reactant 100gm = 1.8 moles Multiply by 3 = 1.8 moles or 180 grams Sodium Hydroxide = 40 gm / total reactant 100 mg = 1.2 moles or 120 grams. 180 grams acetic acid + 120 grams sodium hydroxide = 300 grams. 300 grams divided by 1 liter = 3M So in order to make 3 M sodium acetate combine solution, add 180 grams acetic acid and 120 grams sodium hydroxide with 1 liter of water.

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Related questions

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It is 25 moles of Sodium Hydroxide (;

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.

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.

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.

6NaOH + 3I2 = 5NaI + NaIO3 + 3H2O Six moles of sodium hydroxide and three moles of diatomic iodine yield five moles of sodium iodide, one mole of sodium iodate, and three moles of water. Cheers!

0.0349 mol

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.

0.2 mol

The molecular weight of sodium hydroxide is 40g/mol. To get the amount of moles, you have to divide the weight by molecular mass. 12g / 40 is 0.3 moles. This is 300 millimoles.

Sodium reacts with water. 0.652 NaOH moles will form.

0.0214 mol