An answer is not possible without any details.
A neutral solution of about 7 pH.
the equivalence point has been reached
0.0214 mol
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
A neutral solution of about 7 pH.
The number of moles is 1,8.
the equivalence point has been reached
1
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.
This is a titration question: we want to have the same number of hydroxide ions as hydroxide ions so that they will form water and the pH will be neutral. In chemistry, we count atoms and molecules in moles, and we can calculate how many moles of HBr we have, because concentration in molarity is the number of moles divided by the volume in liters... M = moles/V. We plug in what we got: 1.45M = moles/0.0350L, and solve for moles: 0.0508 moles. Now we know we need 0.0508 moles of NaOH, whose molecular weight is 40g/mole. MW x moles = grams, so (40g/mole)(0.0508 moles) = 2.03 g of NaOH.
It is 25 moles of Sodium Hydroxide (;
417.6 grams BaOH (1mol BaOH/137.3g ) = 3.042 mol
MgCl2(aq) + 2KOH(aq) --> Mg(OH)2(s) + 2KCl(aq)It is the molar ratio in the equation. Every mole of magnesium chloride requires 2 moles of potassium hydroxide. Thus 3 moles would need 6 moles of alkali for complete reaction. We don't have that much, so potassium hydroxide is the limiting reactant and we can only use 2 moles of the magnesium chloride and produce 2 moles of magnesium hydroxide.
0.1 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.
0.0349 mol