117.22 g are needed to react with an excess of Fe2O3 to produce 156.2 g of Fe.
Explanation:
Moles of Fe = Mass of Fe in grams / Atomic weight of Fe
= 156.2 / 55.847
Moles of Fe = 2.79.
The ratio between CO and Fe id 3 : 2.
Moles CO needed = 2.79 * (3 / 2)
= 4.185.
To calculate Atomic weight of CO,
Atomic weight of carbon = 12.011
Atomic weight of oxygen= 15.9994
Atomic weight of CO = 12.011 + 15.9994 = 28.01 g / mol.
Mass of CO = 4.185 * 28.01 = 117.22 g.
75.0 mL in liters:
75.0 / 1000 => 0.075 L
1 mole -------------------- 22.4 L ( at STP)
( moles Hg) ------------- 0.075 L
moles Hg = 0.075 x 1 / 22.4
moles = 0.075 / 22.4
= 0.00334 moles of Hg
Hg => 200.59 u
1 mole Hg ----------------- 200.59 g
<span>0.00334 moles Hg ----- ( mass Hg )
</span>
mass Hg = 200.59 x 0.00334 / 1
mass Hg = 0.6699 / 1
= 0.6699 g of Hg
Answer:
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to solve this problem by firstly setting up the equilibrium expression for the given reaction, in agreement to the law of mass action:
Next, we plug in the given concentrations on the data table to obtain:
Regards!
Answer:
The answer to your question is: 6 moles of HNO₃
Explanation:
Data
Volume = 25 ml
Concentration = 6 M HNO₃
Diluted 100 ml
Formula
Molarity = # moles / volume
# of moles = Volume x Molarity
Process
# of moles = 0.10 x 6
= 6 moles