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
Sodium reacts with water. 0.652 NaOH moles will form.
Oxalic acid forms an oxalate salt when reacted with two equivalents of base - the name of the salt depends on the composition of the base. For example, if one mole of oxalic acid reacts with two moles of sodium hydroxide (NaOH), then one mole of sodium oxalate and two moles of water are formed. (Sodium oxalate is Na2C2O4.) If one mole of oxalic acid reacted with two moles of ammonia (NH3), then one mole of ammonium oxalate ((NH4)2C2O4) and two moles of water are formed.
8 g NaOH x 1 mole NaOH/40 g = 0.2 moles NaOH
400 grams of nickel sulphate (anhydrous) is equivalent to 2,58 moles.
1
5,25 moles (in anhydrous sodium sulphate)
It is 25 moles of Sodium Hydroxide (;
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.
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!
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.
6.066 into 10(up 23 )
60 g NaOH x 1 mole NaOH/40 g NaOH = 1.5 moles NaOH