Okay... what are the following
Answer:
88.24
Explanation:
find the total mass of the compound using there RAM which has been give then the mass of caborn
Two precursor alkenes
H₃C CH₃
I I
H₂C=C-CH-CH₃ 2,3-dimethyl-1-butene
H₃C CH₃
I I
H₃C-CH=CH-CH₃ 2,3-dimethyl-2-butene
alkane
H₃C CH₃
I I
H₃C-CH-CH-CH₃ 2,3-dimethylbutane
H₃C CH₃ H₃C CH₃
I I I I
H₂C=C-CH-CH₃ + H₂ → H₃C-CH-CH-CH₃
H₃C CH₃ H₃C CH₃
I I I I
H₂C-C=CH-CH₃ + H₂ → H₃C-CH-CH-CH₃
Answer:
The answer to your question is 3 moles of AlCl₃
Explanation:
Process
1.- Write and balance the equation
Al(NO₃)₃ + 3NaCl ⇒ 3NaNO₃ + AlCl₃
2.- Determine the limiting reactant
Theoretical proportion 1 mol Al(NO₃)₃ : 3 moles of NaCl
Experimental proportion 4 moles Al(NO₃)₃ : 9 moles NaCl
From these values, we determine that the limiting reactant is NaCl because the number of moles increases three times and the number of moles of Al(NO₃)₃ increases four times.
3.- Determine the amount of AlCl₃ using proportions
3 moles of NaCl --------------- 1 mol of AlCl₃
9 moles of NaCl ---------------- x
x = (9 x 1) / 3
x = 9 /3
x = 3 moles
