how many moles of iodomethanel are in 4dm cube of a2M aqueos solution?

Question

given that 4g of methane were burnt in 7dm cube oxigen what is the limiting reagent?
if 5dm cube of CO2 were produced what is the yield of the reaction

Answer 1

CH2OHCl: 66,45g/mol
Do you mean 4dm^3 or a cube with 4dm length edges?
If you mean 4dm^3 then
4dm^3= 4*10^3 cm^3 = 4*10^3 mL= 4L
2M= ?/4L
?= 8 moles of CH2OHCl
if you meant 4dm edges then
4dm= 4*10^1cm
(4*10^1cm)^3= 6*10^4cm^3 (actually it is 6,4 but because of the lack of figures you must report 6 instead of 6,4)
6*10^4cm^3= 6*10^4 mL = 6*10^1 L
Therefore
2M= ?/6*10^1L
?= 1,2*10^2 moles of CH2OHCl

Part II
CH4 + 2O2 → CO2 + 2H2O
7dm of O (again: 7dm edges or 7dm^3???)
for 7dm sides
(7*10^1cm)^3= 3*10^2 L
O density = 1,43 g/L
Therefore: 5*10^2g or 0,5 kg of Oxygen
4g CH4*(1mol CH4/16g CH4)*(1mol CO2/1mol CH4)= 1*10^1g of CO2
If you compare the amount of Oxygen and methane and the molar ratios there is no need to do the same for O. Obviously CH4 is the limiting reagent
CO2 density: 1,98*10^-3 g/mL
1,98kg/m^3
If 5dm (again the same problem) were produced, then
for 5dm edges: 2*10^2g of CO2 (obviously not this one)
for 5dm^3: 5*10^3 cm^3 of CO2: 1*10^1g of CO2 this is it!)
Now the yield
If you would have added more figures we would have had
(9,9g/11g)*100= 90% yield
But because of the lack of figures we end up with
(1*10^1g /1*10^1g)*100= 1*10^2 % (you got the maximum yield)

Answer 2

4dm cube = (40cm)^3 = 64000cc = 64 L
2Mol/L x 64L = 128 mol
Assuming standard conditions.
11.2 dm cube of O2 is required, therefore O2 limits

1 unit of CO2 is produced for each 2 units of oxygen used. Theoretical CO2 = 5.6 L 5x100/5.6 = 89% (90% sig fig)

Answer 3

You don't need complicated maths or formulae.

In one dm3 of 1Molar ANYTHING there's 1 mole
so in 4dm3 of 2Molar solution there must be 8 moles

If you want to know the MASS of iodomethane in the solution its the relative molecular mass times the number of moles
142 x 8 = 1136g ( Iodomethane CH3I = (C) 12 + (H3) 3 + (I) 127 = 142 )