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dolphi86 [110]
4 years ago
12

Consider the following reaction.Cr2O3(s) + 3CCl4(l) 2CrCl3(s) + 3COCl2(g). When the green solid is mixed with the colorless liq

uid, the mixture starts to bubble and fume. When all action has stopped, a dry purple solid containing solid green specks remains. Which substance is the limiting reactant?A) Cr2O3B) CCl4C) CrCl3D) COCl2E) there is no limiting reactant
Chemistry
1 answer:
Elena L [17]4 years ago
6 0

Answer:

The answer to your question is: letter B

Explanation:

Reaction

                 Cr2O3(s)   +   3CCl4(l)   ⇒  2CrCl3(s)  +   3COCl2(g)

From the information given and the reaction, we can conclude that:

Green solid = Cr2O3 (s)     "s" means solid

Colorless liquid = CCl4 (l)    "l" means liquid   and is the other reactant

Purple solid = CrCl3(s)        CrCl3 is purple and "s" solid

Then, as a green specks remains it means that the excess reactant is Cr2O3, so, CCl4 is the limiting reactant.

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Changing subscripts is one of the steps of balancing a chemical equation.<br> True or False
Lostsunrise [7]

Answer:

False

Explanation:

Changing the coefficients is one of the steps of balancing a chemical equation. Changing the subscript changes the compounds being used, while changing the coefficient changes the amount of each compound being used.

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Jobisdone [24]
ΔHrxn = ΣδΗ(bond breaking) - ΣδΗ(bond making)

Bond enthalpies,
N ≡ N ⇒ 945 kJ mol⁻¹
N - Cl ⇒ 192 kJ mol⁻¹
Cl - Cl⇒ 242 kJ mol⁻¹

According to the balanced equation,
ΣδΗ(bond breaking)  = N ≡ N x 1 + Cl - Cl x 3
                                  = 945 + 3(242)
                                  = 1671 kJ mol⁻¹
ΣδΗ(bond making)    = N - Cl x 3 x 2
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δHrxn = ΣδΗ(bond breaking) - ΣδΗ(bond making)
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4 years ago
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Calculate the masses of oxygen and nitrogen that are dissolved in of aqueous solution in equilibrium with air at 25 °C and 760 T
shtirl [24]

Explanation:

Let us assume that the volume of given aqueous solution is 7.5 L.

Therefore, according to Henry's law, the relation between concentration and pressure is as follows.

                C = \frac{P}{K_{h}}

where,   pressure (P) = 760 torr = 1 atm

According to Henry's law, constants for gases in water at 25^{o}C are as follows.

  p(O_{2}) = 0.21 atm = 0.21 bar

   p(N_{2}) = 0.78 atm = 0.78 bar

   K_{h} for O_{2} = 7.9 \times 10^{2} bar/mol

    K_{h} for N_{2} = 1.6 \times 10^{3} bar /mol

Since, 21% oxygen is present in air so, its mass will be 0.21 g. Similarly, 78% nitrogen means the mass of nitrogen is 0.78 g.

Therefore, concebtrations will be calculated as follows.

      C(O_{2}) = \frac{0.21}{7.9 \times 10^{2}} = 2.66 \times 10^-4 mol/L  

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Now, we will calculate the number of moles as follows.

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Hence, mass of oxygen will be as follows.

         Mass of O_2 = 32 \times 1.995 \times 10^-3 \times 1000

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