The mole of a metal, such as sodium, is generally assumed to consist of Avogadro's Number of single atoms. The gram Atomic Mass of sodium is 22.9898. The number of moles in 120 grams of Na is therefore 120/22.9898, or 5.22, rounded to the justified number of significant digits.
The number of moles of NaOH in the solution will be equal to the molarity (M, the concentration) times the volume in Liters. For this problem, the volume is .120 Liters.
Molar mass NaOH:
Na = 23
O = 16
H = 1+
------
40g/Mol 120/40 = 3
The number of atoms is 0,72265690284.10e23.
1
It is 25 moles of Sodium Hydroxide (;
The molecular weight of sodium hydroxide is 40. Therefore 0.150 moles would be 40 x 0.15 = 6g.
0,15 moles NaOH contain 6 g.
For this you need the atomic (molecular) mass of NaOH. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. NaOH= 40.083.0 grams NaOH / (40.0 grams)= 2.08 moles NaOH
1
It is 25 moles of Sodium Hydroxide (;
0.2 mol
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.
The molecular weight of sodium hydroxide is 40. Therefore 0.150 moles would be 40 x 0.15 = 6g.
60 g NaOH x 1 mole NaOH/40 g NaOH = 1.5 moles NaOH
0,15 moles NaOH contain 6 g.
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.
Need total molar mass sodium hydroxide which you can easily get from your periodic table. 2.2 moles NaOH (39.998 grams/1 mole NaOH) = 88 grams in mass ==============
Molarity = moles of solute/Liters of solution ( 918 ml = 0.918 liters )rearranged algebraically,moles of solute = Liters of solution * Molaritymoles of NaOH = (0.918 l)(0.4922 M)= 0.45184 moles NaOH=======================so,0.45184 moles NaOH (39.998 grams/1 mole NaOH)= 18.1 grams sodium hydroxide needed============================
For this you need the atomic (molecular) mass of NaOH. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. NaOH= 40.083.0 grams NaOH / (40.0 grams)= 2.08 moles NaOH
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.