Your reaction is this...
H3PO4 + 3KOH --> K3PO4 + 3H2O
This type of reaction is called a double replacement.
Note: You didn't put a 4 on the oxygen on the phosphate group on the product side in your question. If it wasnt there the equation would not have been balanced so i put it in myself.
None. All oxidation states stay the same in this reaction.
3koh + h3po4 --> k3po4 + 3h2o ======================== ( site messing up again!! All letters capitalized )
Look at the products. You should already know that one is bound to be water, H2O. You know that the OH comes from KOH, and the H comes from H3PO4. So ask yourself the question: How many OH and how many H are available? You can see that since you have three of one, you will also need three of the other, so you need 3 of the KOH, right? So this is the equation: 3KOH + H3PO4 -> 3H2O + K3PO4
K3PO4 + 3HCl -> 3KCl + H3PO4 Since K on the reactant side has 3 potassium atoms, K on the product side should also have 3 potassium atoms to balance the equation. Since you put the 3 on KCl of the product side, another 3 has to go on the Cl on the reactant side which also matches the 3 hydrogen atoms on the product side in H3PO4. If you check, the equation is now balanced. Everything that appears on the left, equally appears on the right
General Formula for this type of reaction is ACID + BASE ---> SALT + WATER H3PO4 + KOH the base is potassium:K(charge+1) the acid is phosphate(charge-3) -to neutralize the -3 charge you need a +3 charge therefor K must equal 3K the correct formula is K3PO4 the left behind H will form with OH and form H20 H3PO4 + KOH ----> K3PO4 + H2O..... unbalanced H3PO4 + 3KOH ---> K3PO4 + 3H20 ...... BALANCED Above reaction is the net reaction product. Reaction will complete in three steps. 1. First Potassium di hydrogen phosphate will be produced 2. Further neutralization will take place then to form Di Potassium Hydrogen Phosphate 3. Then, finally Potassium Phosphate Tri Basic will be formed. However the solubility of Potassium Di Hydrogen Phosphate in water is far better than of Potassium Phosphate Tri Basic.
The chemical equation is written out as:3KOH + H3PO4 -> 3H2O + K3PO4.So, if you do your mole bridge, you take your moles of Phosphoric acid (1.56) and do some quick stoichiometry.1.56 mole H3PO4 x (3 mole KOH)/(1 mol H3PO4). This comes out to.... 4.68 moles of KOH is required to neutralize 1.56 moles of H3PO4.
The balanced Saponification equation for trimyristin and KOH is as follows: C45H86O6 + 3KOH -----> C3H5(OH)3 + 3K+-C14H27O2 This is a very condensed formula... If I were you, I would suggest looking up the structural formula for trimyristin, which can be found at the link below, and then do the replacement reaction.
The possible reaction products are aluminum chloride and nitric acid. Going from a strong acid to a weak acid will drive a reaction, but both HCl and nitric acid are strong acids. Producing a precipitate will drive a reaction, but aluminum chloride is reasonably soluble. In short, there's no obvious reason for a metathesis reaction.
A balanced equation for the preparation of soap from triacylglycerol is (C18H29O2)3-C3H5O3 + 3KOH -> 3C18H29O2K + HOCH2CH(OH)CH2OH. The acyl portions are all derived from linolenic acid and use potassium hydroxide as the base.
i think you are misunderstanding the concept of net ionic equation. Iron III clhloride potassium hydroxide does not have a net ionic equation UNLESS you mean the reaction between those two
Al2(SO4)3 + 6KOH --> 2 [Al(OH)3](s) + 3K2SO4or in excess hydroxide:Al2(SO4)3 + 8KOH --> 2K+ + 2 [Al(OH)4]-(aq) + 3K2SO4
Ammonia gas has characteristic properties which are used to identify it.Some of Physical properties are:Color: colorlessodor: characteristic strong pungent smell(ammoniacal smell )Physiological nature: causes an irritating, burning sensation in upper part of nasal passage, fatal in large doses.Density: Lighter than air (V.D of ammonia = 8.5)Melting point: -77.70CBoiling point: -33.40CBut physical properties can't be relied to identify a substance, therefore chemical tests should be used.Chemical tests:1. A glass rod dipped in conc. HCl acid brought near ammonia gas gives dense white fumes. This is due to formation of Ammonium ChlorideNH3 + HCl ---> NH4Cl2. When ammonia is passed through Copper sulfate solution a pale blue precipitate of copper hydroxide is formed.3.Colorless Nessler's Reagent turns pale brown and on passing excess ammonia gives a brown precipitate.2K2[HgI4] + NH3 + 3KOH ---> H2N.HgO.HgI + 7KI + 2H2O(nessler's reagent) (Brown ppt.)4.All salts of ammonia on heating with alkali produce ammonia gas which can be identified