The gram molecular formula unit mass of sodium hydroxide is 40.00. Therefore,4.75 g constitutes 4.75/40.00 or 1.19 X 10-1 mole, to the justified number of significant digits.
1
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.
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.
I need to see the balanced equation to work!NaOH + HCl --> NaCl + H2O ( good, all one to one )Now, find molarity HCl ( sodium, or sodium hydroxide; no matter )(17.65 mL)(0.110 M NaOH) = (25.00 mL)(X M HCl)= 0.07766 M HCl-------------------------now,Molarity = moles of solute/Liters of solution ( 25.00 mL = 0.025 L)0.07766 M HCl = X moles/0.025 Liters= 0.001942 moles HCl---------------------------------------formal set up, though not needed0.001942 moles HCl (1 mole NaOH/1 mole HCl)= 0.00194 moles sodium hydroxide used=============================
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.
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.
60 g NaOH x 1 mole NaOH/40 g NaOH = 1.5 moles NaOH
20 moles of NaOH needed to neutralize 20 moles of nitric acid
0,15 moles NaOH contain 6 g.
8 g NaOH x 1 mole NaOH/40 g = 0.2 moles NaOH
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.