Moles= molarity x liter
=0.2500 M x 0.500 L
= 0.1250 mol
mass C2H2O4 . 2H2O = 0.1250 mol x (126.068 g / 1 mol)
=15.76 g
To determine the molar mass of the unknown gas, we use Graham's Law of Effusion where it relates the effusion rates of two gases with their molar masses. It is expressed as r1/r2 = √M2/M1. We calculate as follows:
Let 1 = argon gas 2 = unknown gas
r2 = 0.91r1r1/r2 = 1/0.91
1/0.91 = √M2/M1 = √M2/40M2 = 48.30 g/mol
Answer:
The answer to your question is letter b) X
Explanation:
Data
W = 5 units
X = 6 units
Chemical reaction
2W + 3X ⇒ 3Y + Z
To determine which reactant is the limiting reactant, we must use proportions:
-Theoretical proportion W / X = 2 / 3 = 0.67
- Experimental proportion W / X = 5 / 6 = 0.83
As the experimental proportion is higher than the theoretical proportion, we conclude that the amount of W is higher in the experiment so the limiting reactant is X.
