Answer:
The general preparation of esters( for example ethyl ethanoate) is through a process known as ESTERIFICATION.
Explanation:
The formation of an ester by the reaction between an alkanol and an acid is known as esterification. This reaction is extremely slow and reversible at room temperature, and is catalyzed by a high concentration of hydrogen ions.
In the preparation of one of the simpler esters known as ETHYL ETHANOATE the reactants include ethanol(an alcohol) and glacial ethanoic acid(a carboxylic acid) in the presence of concentrated tetraoxosulphate VI acid as a CATALYST. Note that, a catalyst is any substance that is able to increase the rate of a chemical reaction.
The mixture is warmed in a water bath( hot but not boiling) for about 25 minutes. The mixture is poured into a beaker partially filled with a sodium or calcium chloride to remove interacted ethanol. The ethyl ETHANOATE floats on the mixture as oily globules.
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1. The Alpha particle is a helium nucleus, is positively charged & would barely penetrate a piece of paper.<span />
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Answer:
30.4 g. NH3
Explanation:
This problem tells us that the hydrogen (H2) is the limiting reactant, as there is "an excess of nitrogen." Using stoichiometry (the relationship between the various species of the equation), we can see that for every 3 moles of H2 consumed, 2 moles of NH3 are produced.
But before we can use that relationship to find the number of grams of ammonia produced, we need to convert the given grams of hydrogen into moles:
5.4 g x [1 mol H2/(1.008x2 g.)] = 2.67857 mol H2 (not using significant figures yet; want to be as accurate as possible)
Now, we can use the relationship between H2 and NH3.
2.67857 mol H2 x (2 mol NH3/3 mol H2) = 1.7857 mol NH3
Now, we have the number of moles of ammonia produced, but the answer asks us for grams. Use the molar mass of ammonia to convert.
1.7857 mol NH3 x 17.034 g. NH3/mol NH3 = 30.4 g. NH3 (used a default # of 3 sig figs)