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60 g NaOH x 1 mole NaOH/40 g NaOH = 1.5 moles NaOH
1. Three moles of sodium contain 18,06642387.1023 atoms. 2. The mass of three moles of sodium is 68,97 grams.
20 moles of NaOH needed to neutralize 20 moles of nitric acid
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.
The answer is approx. 2 moles (for anhydrous sodium sulfate).
1
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.
0.2 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.
There are 10 moles present in 585 g of sodium chloride.
How many MOLES of sodium nitrate are present in 2.85 grams of this compound ?
Sodium reacts with water. 0.652 NaOH moles will form.
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.
The molecular weight of sodium hydroxide is 40. Therefore 0.150 moles would be 40 x 0.15 = 6g.
1. Three moles of sodium contain 18,06642387.1023 atoms. 2. The mass of three moles of sodium is 68,97 grams.
60 g NaOH x 1 mole NaOH/40 g NaOH = 1.5 moles NaOH
1. Three moles of sodium contain 18,06642387.1023 atoms. 2. The mass of three moles of sodium is 68,97 grams.