Answer: 631.8 g
Explanation:

It can be seen from the balanced chemical equation, 2 moles of ethane reacts with 7 moles of Oxygen gas to produce 4 moles of carbon dioxide and 6 moles of water.
Ethane is the limiting reagent as it limits the formation of product.
Thus, if 2 moles of ethane produce 6 moles of water.
11.7 moles moles of ethane produce=
of water.
Mass of water= no of moles
Molar mass
Mass of water= 35.1
18g/mol= 631.8 g
Answer:
146 g
Explanation:
Step 1. Calculate the <em>molar mass</em> of NaNO₃
Na = 22.99
N = 14.01
3O = 3 × 16.00 = 48.00
Total = 85.00 g/mol
Step 2. Calculate the <em>mass</em> of NaNO₃
Mass of NaNO₃ = 1.72 × 85.00/1
Mass of NaNO₃ = 146 g
Answer:
C₂ = 0.149 M
Explanation:
Given data:
Initial concentration = 0.407 M
Initial volume = 2.56 L
Final volume = 7.005 L
Final concentration = ?
Solution:
Formula:
C₁V₁ = C₂V₂
C₁ = Initial concentration
V₁ = Initial volume
C₂ = Final concentration
V₂ =Final volume
Now we will put the values.
0.407 M × 2.56 L = C₂ × 7.005 L
1.042 = C₂ × 7.005 L
C₂ = 1.042 M.L / 7.005 L
C₂ = 0.149 M
Answer:
<span>Chlorine (Cl) is the oxidizing agent because it gains an electron.
Explanation:
Reaction is as follow,
</span><span> Cl</span>₂<span> (aq) + 2 Br</span>⁻<span> (aq) </span>→ <span> 2Cl(aq) + Br</span>₂ <span>(aq)
Oxidation Reaction:
2 Br</span>⁻ → Br₂ + 2 e⁻
Two atoms of Br⁻ (Bromide) looses two electrons to form Br₂ molecule. Hence it is oxidized and is acting as reducing agent.
Reduction Reaction:
Cl₂ + 2 e⁻ → 2 Cl⁻
One molecule of Cl₂ gains two electrons to form two chloride ions (Cl⁻). Therefore, it is reduced and has oxidized Br⁻, Hence, acting as a oxidizing agent.
Answer:
338.00 mL
Explanation:
The lead ions come from the salt Pb(NO₃)₂ and the iodide from the acid HI, so the balanced reaction is:
Pb(NO₃)₂(aq) + 2HI(aq) → PbI₂(s) + 2HNO₃(aq)
So, the stoichiometry is 1 mol of Pb(NO₃)₂ to 2 moles of HI, then:
1 mol of Pb(NO₃)------------------------------------ 2 moles of HI
0.600 mol of Pb(NO₃)₂--------------------------- x
By a simple direct three rule:
x = 1.200 mol of HI
The acid has concentration equal to 3.550 mol/L, the volume (V) is the number of moles divided by the molar concentration:
V = 1.200/3.550 = 0.338 L
V = 338.00 mL