Answer:
The glassware and solvent for the production of grignard reagent and its reaction should by dry and anhydrous so as to prevent the conversion of the grignard reagent into saturated alkane. Acetone, is not a good solvent for the generation of grignard reagent because it has the potential of forming alcohol.
Explanation:
Grignard reagent is a compound formed by the reaction of halide of alkyl or alkene with magnesium metal. This makes the compound more nucleophilic. Thus the availability of an electrophile around it will evoke a quick reaction. The presence of water in an acidic condition will generate hydroxonium ions which are highly electrophilic. There will therefore be an introduction of hydrogen ions to the grignard reagent and which will displace the MgBr leading to the formation of alkane.
Acetone tends to produce alcohol when they are exposed to grignard reagent, thus they are not appropriate to be used as solvent. This is due to the resultant highly electrophilic nature of the carbonyl carbon on the acetone, thus will react with the nucleophilic carbon on the grignard. The reaction can be represented as follows:
RMgBr + CH3(CO)CH3 + H (with hydrogen ions) >>>RCOH + Mg(OH)Br
Ok so the way I do it is as simple as possible.
Firstly look at the reactants and products ( there can be one reactant and one product or more ) you will usually be given the moles of the reactant or products, if you are given grams you can convert into moles by this convertion ( grams/R.M.M ) where R.M.M is the relative atomic mass of your substance ( the mass number of all of the elements in your substance).
Ok when you have moles now look at the ratio between the products and reactants. Usually you will won't know the moles of one substance therefore you will be asked to find moles or mass of that substance.
For example:
When 16 grams of oxygen and 1 gram of hydrogen gas react to produce water. Find the number of grams of water being produced.
O2 + 2H2 -> 2H2O
16g 2g xg
Here we're told the mass of the reactants. In stoichiometry we need to work with moles therefore you need to calculate moles of the reactants.
Firstly find the R.M.M of each reactant.
R.M.M of O2 is 16+16=32 since it's diatomic we add atomic masses of two oxygen atoms.
R.M.M of H2 is 1+1=2, it's also diatomic. (Diatomic two atoms of the same element are joined together). (Ignore the number 2 in front of H2, this number shows us the ratio relationship between reactans or products, i.e when we balance an equation.)
Ok so now find moles:
We have 16 grams of O2
16/R.M.M
16/32 = 0.5 moles
We have 2 grams of H2
1/R.M.M
2/2 = 1 mole
Now back to the equation.
O2 + 2H2 -> 2H2O
0.5 moles 1mole xmoles (it's xmoles because we don't know molarity of water that's what we have find firstly in order to find grams.)
Now look at the ratio between any reactant and product i.e you can choose which reactant to compare to the product, it doesn't make a different ( I will do two or you can do two at the same time)
1st method:
Look at the ratio between O2 and H2O from the reaction above we see the ratio is 1:2 therefore for every 0.5 moles of O2 you get 1 mole of H2O.
1:2
0.5 : x
0.5*2 = 1
2nd method;
Look at the ratio betweem H2 and H2O from the reaction above we see the ratio is 2:2 or 1:1. We have 1 mole of H2 there we must have 1 mole of H2O. We see this is true as both methods give us 1 mole of H2O.
3rd method ( combined):
Look at the ratio between O2, H2 and H2O.
We see that the ratio is 1:2:2
So we have 0.5:1:x
If we multiply 0.5 *2 it equals 1 mole
If we multiply 1*1 we get 1 moles.
Any method is correct and it's up to you to find a comfortable way.
We're not finished in the question we are asked for the mass of water.
So just multiply the number of moles (1mole) by R.M.M of H2O.
1 * R.M.M
R.M.M of H2O = 1+1+16=18
1*18= 18 grams.
And you're finished.
I am sorry if this is so long I want you to understand as much as possible.
In stoichiometry you can also be asked about the empirical formula of a substance. I can show you how do it. If you have any question just tell me.
Hope this helps :).
Answer:
-7.34 kilo Joules is the change in enthaply when 20.0 grams of nitrogen triiodide decomposes.
Explanation:
Mass of nitrogen triiodide = 20.0 g
Moles of nitrogen triiodide = 

According to reaction, 2 moles of nitrogen triiodide gives 290.0 kilo Joules of heat on decomposition ,then 0.05063 moles of nitrogen triiodide will give :

-7.34 kilo Joules is the change in enthaply when 20.0 grams of nitrogen triiodide decomposes.