4P + 502 -> P4O10 this is the answer
It would be the same amount. So, 45 ml of NaOH is required to be added to the 45 ml of HCI to neutralize the acid fully. Here is a brief calculation:
Firstly, here is your formula: M(HCI) x V(HCI) = M(NaOh) x V(NaOH)
With the values put in: 0.35 x 45 = 0.35 x V(NaOH)
= 45 ml.
There is 45 ml of V(NaOH)
Let me know if you need anything else. :)
- Dotz
Using the law of dilution :
Mi x Vi = Mf x Vf
2.00 x Vi = 0.15 x 100.0
2.00 x Vi = 15
Vi = 15 / 2.00
Vi = 7.5 mL
hope this helps!
Answer:
33.3 g AlCl3
Explanation:
First:
You need a balanced chem equation.
2Al + 3Cl2 --->2AlCl3
So now you use this to set up train track method which helps us cancel out the units. Also we dont care about chlorine because it is excess.
6.73g Al x 1mol Al/26.98g Al x 2mol AlCl3/2molAl x 133.34g AlCl3/1molAlCl3
= 33.3 g AlCl3
