<u>Answer:</u> Amount of iron produced is 101.36 grams.
<u>Explanation:</u>
For the reaction of aluminium and iron oxide, the equation follows:
By Stoichiometry of the reaction,
When 1 mole of aluminium oxide is produced, then 2 moles of iron is also produced.
So, when 0.905 moles of aluminium oxide is produced, then 
of iron is also produced.
Now, to calculate the amount of iron produced, we use the equation:
Molar mass of iron = 56 g/mol
Moles of iron = 1.81 moles
Putting values in above equation, we get:
Hence, amount of iron produced is 101.36 grams.
Answer:
pCH4 = 0.9184 atm
pCCl4 = 0.9184 atm
pCH2Cl2 = 0.2832 atm
Explanation:
Step 1: Data given
The equilibrium constant, Kp= 9.52 * 10^-2
Temperature = 350 K
Each have an initial pressure of 1.06 atm
Step 2: The balanced equation
CH4(g) + CCl4(g) ⇆ 2CH2Cl2(g)
Step 3: The pressure at the equilibrium
pCH4 = 1.06 - X atm
pCCl4 = 1.06 - X atm
pCH2Cl2 = 2X
Step 4: Calculate Kp
Kp = (2X)² / (1.06 - X)*(1.06 - X)
9.52 * 10^-2 = 4X² / (1.06 - X)*(1.06 - X)
X = 0.1416
Step 5: Calculate the partial pressure
pCH4 = 1.06 - 0.1416 = 0.9184 atm
pCCl4 = 1.06 - 0.1416 = 0.9184 atm
pCH2Cl2 = 2 * 0.1416 = 0.2832 atm
Kp = (0.2832²) / (0.9184*0.9184)
Kp = 9.52 * 10^-2
pCH4 = 0.9184 atm
pCCl4 = 0.9184 atm
pCH2Cl2 = 0.2832 atm
<span>Li ----> Li+ + e-(oxidation)
now comes </span>
<span>Reduction: </span>
<span>H2 + 2e- ------> 2H
now by
</span>Multiplying (1) by 2 and adding in (2) will give you the required equation..
..
<span>2Li + H2 ---------> 2LiH
</span>half reaction are either oxidation or reduction
Answer:
160 K, using 2 SF,
Explanation:
Ideal Gas law, rearrange to solve for T (K), answer about 158.3 round to 2SF
PV = nRT
