The molar mass of the unknown compound is calculated as follows
let the unknown gas be represented by letter Y
Rate of C2F4/ rate of Y = sqrt of molar mass of gas Y/ molar mass of C2F4
= (4.6 x10^-6/ 5.8 x10^-6) = sqrt of Y/ 100
remove the square root sign by squaring in both side
(4.6 x 10^-6 / 5.8 x10^-6)^2 = Y/100
= 0.629 =Y/100
multiply both side by 100
Y= 62.9 is the molar mass of unknown gas
The limiting reactant is the chocolate chips (they would run out first) because you can only make 3 batches with 300 chips.
Communication will be the answer
The balanced equation for the reaction between Mg and O₂ is as follows
2Mg + O₂ --> 2MgO
stoichiometry between Mg and O₂ is 2:1
number of Mg reacted - 4.00 mol
if 2 mol of Mg reacts with 1 mol of O₂
then 4.00 mol of Mg requires - 1/2 x 4.00 = 2.00 mol of O₂
then the mass of O₂ required - 2.00 mol x 32.0 g/mol = 64.0 g
64.0 g of O₂ is required for the reaction
