I need help (1 Viewer)

[atrocityx]

man, i'm asking too many questions and making new threads ... =.=" ... i think im invading the chemistry thread.. SORRRYYY DDD= i will stop after exams~ BUT ANYWAYS. I REALLY NEED HELP WITH THESE QUESTION CAUSE I HAVE AN EXAM ON THIS WEEK T.T" AND THEY'RE REALLY KILLING ME .

1) water of crystallisation in a compound can be removed by heating the compound in a crucible. determine the perecentage of the water crystallisation in a compound using the following experimental results : mass of empty crucible: 23.64g, mass of crucible+compound: 27.72g; mass of crucible + compound after heating: 26.56g ??

2) half equations showing the ionisation of the following elements
i) k -> (would it be k+ + e? =S)
ii) al -> ?

3) 39.1g of potassium reacts with excess oxygen to produce 47.1g of potassium oxide. calculate the empirical formula of the oxide. ( would you minus 47.1 - 39.1 to find the oxygen then find the empirical formula from that? " )

4) if 32.0g of methane and 2.0 moles of oxygen are reacted, not all of the gases are used up. determine which one will be left over and the mass of carbon dioxide produced.


OMG IM SOO CONFUSED T_T"

 

[Steth0scope]

1. mass of compound before heating = (mass of crucible + mass of compound) - mass of crucible = 27.72 - 23.64 = 4.08g
mass of compound after heating = (mass of crucible + mass of compound after heating) - mass of crucible = 26.56 - 23.64 = 2.92g

mass of compound lost = mass of crystallised water (from info) = mass of compound before heating - mass of compound after heating = 4.08 - 2.92 = 1.16g. Therefore % crystalised = 1.16g / 4.08 x 100 = ...?

(did calcs in head .. check them with calculator)

2. k ---> k+ + e-
al ---> al3+ + 3e-

3. Need a paper and pen to do it.

4. First of all need to balance reaction: CH4 + 2O2 --> CO2 + 2H2O
Moles of CH4 available = (32g) / (16g/mol) = 2mol
Moles of O2 available = 2mol
But from reaction need twice as much O2 as CH4 i.e. CH4 is in excess and only 1 mol of it will be used. Thus, 1 mol of CO2 is also formed.

Mass of CO2 = molar mass x moles = (44g/mol) x 1 mol = 44g.

 

[Xcelz]

3) 39.1g of potassium reacts with excess oxygen to produce 47.1g of potassium oxide. calculate the empirical formula of the oxide. ( would you minus 47.1 - 39.1 to find the oxygen then find the empirical formula from that? " )
OMG IM SOO CONFUSED T_T"

for this one

equation is
4K</ST1+O2 à 2K2O
so there is 39.1g of K and 47.1g of K2O. therefore there is 8g of O2 </O
<Omole ratio 4:1 > 2</O
<Oso from the 39.1g of K we can deduce that there is 1 mole of K (n=m/M)</O
<Oalso we can deduce there is 0.5 moles of O2 </O
<Ofor the number of moles of K2O --> n=m/M
<On= (47.1)/(2x39.1+8) since 8 = one oxygen atom
<On of K2O = 0.546</O>
<O</O
<Othere is the number of moles for each. work from that with the mole ratio to get your answer.