Q: How many moles of sodium hydroxide are in 40 grams of sodium hydroxide?

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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.

How many MOLES of sodium nitrate are present in 2.85 grams of this compound ?

.73 moles

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.

Related questions

0.2 mol

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.

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

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 ==============

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.

23.3772 grams are there in four tenths moles of sodium chloride

How many MOLES of sodium nitrate are present in 2.85 grams of this compound ?

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============================

.73 moles

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.

55.0 moles of sodium are equal to 1 264,44 g.