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icang [17]
3 years ago
6

Which of the following conversion factors would be used to calculate the number of moles of Cl2 produced if 11 moles of HCl were

present in the following reaction:______
4HCl O2 --> 2Cl2 2H2O
Chemistry
1 answer:
MissTica3 years ago
6 0

<u>Answer:</u> The moles of chlorine gas produced is 5.5 moles

<u>Explanation:</u>

We are given:

Moles of HCl = 11 moles

For the given chemical reaction:

4HCl+O2\rightarrow 2Cl_2+2H_2O

By Stoichiometry of the reaction:

4 moles of HCl produces 2 moles of chlorine gas

So, 11 moles of HCl will be produced from \frac{2}{4}\times 11=5.5mol of chlorine gas

Hence, the moles of chlorine gas produced is 5.5 moles

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Sindrei [870]

Answer: If the gold in the crown was mixed with a less-valuable metal such as bronze of copper then this affects its density by making it to weigh even more and if it weighs more then it has less of chance to float. Observe: Drag each of the crowns into the liquid.

Explanation:

6 0
3 years ago
Determine the [OH−] of a solution that is 0.115 M in CO32−. For carbonic acid (H2CO3), Ka1=4.3×10−7 and Ka2=5.6×10−11.
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Answer:

[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.

Explanation:

Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.

Analysis:

            H₂CO₃(aq)     ⇄     H⁺(aq)    +    HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷

C(i)          0.115M                      0                  0

ΔC              -x                        +x                  +x

C(eq)    0.115M - x                   x                    x

            ≅ 0.115M

Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M

= 4.3 x 10⁻⁷  => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.

In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion  concentration, the hydroxide ion concentration is then calculated from

[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.

________________________________________________________

NOTE: The 2.32 x 10⁻⁴M  value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.

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