Answer:
Esterification reaction
Explanation:
An esterification reaction is an organic reaction involving an organic acid and an alkanol to give an ester or an ethanoate and water
Like the name suggests, an ester is the product formed in an esterification reaction alongside water. It is like a neutralization reaction but this time it solely contains organic molecules. These molecules react with each other to give rise to another organic molecule which is a member of a different homologous series.
Practically, to form ethyl ethanoate, ethanoic acid react with ethanol in the presence of concentrated sulphuric acid which catalyses the reaction.
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In which state of matter is there no particle movement
Solid is the state of matter that has no particle movement. Solid is tightly packed, rigid, and does not takes the shape of the container. The only movement that they have is the vibration movement, otherwise, it is still and has no particle movement.
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Phosphorus is in group 15 meaning it have 5 valence electrons. This means that it needs 3 more electrons to create a full outer shell. As these three electrons are negatively charged it means that P is a 3- ion (it’s an anion [negatively charged ion])
Y=15 x=60
•1/3 •1/3
y=5 x=45
4.20 mol Al would react completely with 4.20 x (1/2) = 2.10 mol Fe2O3, but there is not that much Fe2O3 present, so Fe2O3 is the limiting reactant. (1.75 mol Fe2O3) x (2/1) x ( 55.8452 g Fe/mol) = 195 g Fe 3 MgO + 2 H3PO4 → Mg3(PO4)2 + 3 H2O (15.0 g MgO) / (40.3045 g MgO/mol) = 0.37217 mol MgO (18.5 g H3PO4) / (97.9953 g H3PO4/mol) = 0.18878 mol H3PO4 0.18878 mol H3PO4 would react completely with 0.18878 x (3/2) = 0.28317 mole of MgO, but there is more MgO present than that, so MgO is in excess and H3PO4 is the limiting reactant. Now we must consider why the problem tells us "17.6g of Mg3(PO4)2 is obtained". The first possibility is that it's just there for the sake of confusion -- in which case ignore it and proceed this way: ((0.37217 mol MgO initially) - (0.28317 mole MgO reacted)) x (40.3045 g MgO/mol) = 3.59 g MgO left over However, if the amount of magnesium phosphate obtained is given because the reaction was stopped before it was complete, the amount obtained governs the amount reacted and the amount left over, so proceed this way: (17.6g Mg3(PO4)2) / (262.8581 g Mg3(PO4)2/mol) x (3/1) = 0.20087 mol MgO reacted ((0.37217 mol MgO initially) - (0.20087 mole MgO reacted)) x (40.3045 g MgO/mol) = 6.90 MgO left over
