Answer: 714 g Al2O3
Explanation: Solution attached
First convert mass of O2 to moles
Do the mole ratio between O2 and Al2O3 from the balanced equation.
Convert moles of Al2O3 to mass using its molar mass.
Answer:
2H₂O (liq) + 2e⁻⇒ H₂ (g) + 2OH⁻ (aq)
Explanation:
In reduction-oxidation reaction two reactions take place, one is oxidation and the other is reduction reaction. In an oxidation reaction, there is the loss of an electron whereas in the reduction reaction there is gain of electron occus.
Reduction reaction occurs on the cathode, in a reduction of water there is gain of 2 electrons to gaseous hydrogen in basic aqueous solution. half-reaction for the reduction of liquid water to gaseous hydrogen in basic aqueous solution-
2H₂O (liq) + 2e⁻⇒ H₂ (g) + 2OH⁻ (aq)
Answer:
The mass methane (CH4) required is, 25 grams (option C)
Explanation:
Mass of oxygen gas = 100 g
Molar mass of oxygen gas = 32 g/mole
Molar mass of methane gas = 16 g/mole
<u>Step 1:</u> Balance the reaction
2O2 + CH4 → CO2 + 2H2O
<u>Step 2:</u> Calculate the moles of O2:
Moles O2 = mass of O2 / Molar mass of o2
Moles O2 = 100g / (32g/mole) = 3,125 moles
⇒From the balanced reaction we conclude that 2 moles of O2 react with 1 mole of CH4
⇒ So, 3.125 moles of O2 react with 3.125/2 = 1.563 moles of CH4
<u>Step 3:</u> Calculate the mass of CH4:
Mass of CH4 = moles of CH4 x Molar mass of CH4
Mass of CH4 = 1.563 moles / (16g/ moles) = 25.008 grams CH4
The equilibrium reaction is
2 NOBr ⇄ 2 NO + Br₂
The initial molarity of NOBr is 0.174 mol/1 L = 0.174 M. Let's apply the ICE approach which stands for Initial-Change-Equilibrium.
2 NOBr ⇄ 2 NO + Br₂
I 0.174 0 0
C -2x +2x +x
----------------------------------------------------
E 0.174-2x 2x x
Since E for Br₂ is 1.79*10⁻² M, then that means x = 1.79*10⁻² M. We can compute E for the other compounds.
E for NOBr = 0.174 - 2(1.79*10⁻²) = 0.1382 M
E for NO = 2(1.79*10⁻²) = 0.0358 M
The expression for Kc according to the reaction is:
Kc = [NO]²[Br₂]/[NOBr]²
Kc = [0.0358]²[1.79*10⁻²]/[0.1382]²
<em>Kc = 1.2×10⁻³</em>
Answer:
The correct answer is - 13.33 kJ of heat
Explanation:
To know which one is the limiting reagent, determine the number of moles of each reagent in order .
n(K) = mass/atomic weight = 1.41/39 = 0.036 moles
Density of ICl = Mass/Volume
3.24 = Mass/6.52
Mass of ICl = 21.12 g
n(ICl) = mass/molar mass = 21.12/162.35 = 0.130 moles
2 moles of K reacts with 1 mole of ICl
0.036 moles of K will react with = 0.036/2 = 0.018 moles of ICl
since the amount of moles of ICl is more than 0.018, it is in excess and hence K is the limiting reagent. Now, use the balance equation to determine the amount of heat liberated:
2 moles of K gives out -740.71 kJ of heat
1 mole of K will give out = -740.71/2 = 370.36 kJ of heat
0.036 moles of K will give out = 0.036 × 370.36 = 13.33 kJ of heat
Thus, the correct answer is - 13.33 kJ of heat
