Answer:
The answer is A, if im not mistaken
Answer:
16.6 g of Al are produced in the reaction of 82.4 g of AlCl₃
Explanation:
Let's see the decomposition reaction:
2AlCl₃ → 2Al + 3Cl₂
2 moles of aluminum chloride decompose to 2 moles of solid Al and 3 moles of chlorine gas.
We determine the moles of salt:
82.4 g . 1mol/ 133.34g = 0.618 moles
Ratio is 2:2. 2 moles of salt, can produce 2 moles of Al
Then, 0.618 moles of salt must produce 0.618 moles of Al.
Let's convert the moles to mass → 0.618 mol . 26.98g /mol = 16.6 g
<span>Balancing is making sure there are the same number of atoms on either side of the reaction.
Pb(NO3)2 + Li2SO4--> PbSO4 + LiNO3
There are 2 NO3 groups and 2 Li on the right side, need 2 on the left side.
Need a coefficient of 2 for LiNO3
</span>
<span>write out the balance equation
3NaOh+H3PO4->Na3PO4+3H2O
You are given everything needed to calculate
q=heat transfer=2.2*10^2,
H3PO4 moles= 1.5*10^-3,
NaOH moles=5.0*10^-3
equation is deltaHneutraliztion=q/Moles of limiting reagent
H3PO4 is limiting reagent because lowest moles, and is used up first
Now plug in variables
DeltaH=2.2*10^2(1.5*10^3)= 146.67kj/mole
Notice we had to convert J to kj,</span>