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son4ous [18]
4 years ago
10

1. Interpret the following equation using moles, molecules, and volumes (assume STP). Compare the mass of the reactants to the m

ass of the product. 2N2(g)+3O2(g)--->2N2O3(g)
2. How many moles of chlorine gas will be required with sufficient iron to produce 14 moles of iron (III) chloride?
2Fe(s)+3Cl2(g)--->2FeCl3(g)
Chemistry
2 answers:
Strike441 [17]4 years ago
4 0
1. In the given in this item, we may solve directly the molar masses of the reactants and the products to see if they are matching. 
 Reactant:          (2N2) = (2)(28.0 g/mol) = 56 g
                           (3O2) = (3)(32.0 g/mol) = 96 g
                                                  total = 152 g
 Product:           (2N2O3) = (2)(76 g/mol) = 152 g
The reactant and product have the same masses.

2. From the given balanced chemical equation, we use the stoichiometric ratio,
                         (14 moles of FECl3)(3 moles Cl2/2 moles of FeCl3)
                                        = 21 moles of Cl2
blsea [12.9K]4 years ago
3 0
1. In this reaction, 2 moles of nitrogen gas reacts with 3 moles of oxygen gas to give 2 moles of N2O3 gas. 2 nitrogen molecules react with 3 oxygen molecules to give 2 N2O3 molecules.  Under STP, one mole of an ideal gas occupies a volume of 22.4 liters. So in this reaction, 44.8 liters of nitrogen gas reacts with 67.2 liters of oxygen gas to give 44.8 liters of N2O3 gas.  The total mass of the reactants (N2 and O2) is the same as the total mass of the product (N2O3). This is called mass balance of a chemical reaction.


2. According to the chemical reaction, 3 moles of chlorine gas produces 2 moles of iron(III) chloride.  So, to produce 1 moles of iron(III) chloride, 3/2 (1.5) moles of chlorine gas is required.  Therefore, to produce 14 moles of iron(III) chloride, 14 x 1.5 = 21 moles of chlorine is needed.
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Katen [24]

Answer:

Keq'>1\\\Delta G'

Explanation:

Hello,

In this case, for the given reaction, the equilibrium constant turns out:

Keq=\frac{[B]}{[A]}=\frac{0.5M}{1.5M} =1/3

Nonetheless, we are asked for the reverse equilibrium constant that is:

Keq'=\frac{1}{Keq}=3

Which is greater than one.

In such a way, the Gibbs free energy turns out:

\Delta G'=-RTln(Keq')\\

Now, since the reverse equilibrium constant is greater than zero its natural logarithm is positive, therefore with the initial minus, the Gibbs free energy is less than zero, that is, negative.

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3 years ago
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3 years ago
Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation: N2 1 g 2 1 3H2 1 g
bekas [8.4K]

Answer:

Mass of NH3 produced = 1217 g or 1.217*10^3 g

Explanation:

<u>Given:</u>

Mass of N2 = 1.003*10^3 g

Mass of H2 = 5.003*10^2 g

<u>To determine:</u>

Maximum mass of NH3 that can be produced when the given amounts of N2 and H2 combine

<u>Calculation:</u>

The chemical reaction corresponding to the production of ammonia is:

N2(g)+3H2(g)\rightarrow 2NH3(g)

Based on the reaction stoichiometry:

1 mole of N2 combines with 3 moles of H2 to form 2 moles of NH3

moles\ of\ N2 = \frac{Mass\ N2}{Molar\ mass N2} = \frac{1.003*10^{3}g }{28g/mol} =35.8\ moles

moles\ of\ H2 = \frac{Mass\ H2}{Molar\ mass H2} = \frac{5.003*10^{2}g }{2g/mol} =250\ moles

Since the moles of N2 is less than that of H2, the limiting reagent will be N2 which would in turn determine the amount of NH3 formed.

Based on the reaction stoichiometry the N2 : NH3 ratio = 1:2

Therefore,

moles\ of\ NH3\ produced = 2*35.8 = 71.6\ moles\\\\Mass\ of\ NH3\ produced = moles*molar mass = 71.6\ moles*17\ g/mol = 1217 g

8 0
3 years ago
Which atom has a change of oxidation number of +4 in the pictured redox reaction?
SpyIntel [72]

Answer:

The answer to your question is the letter d. S

Explanation:

Data

Change of +4 in the oxidation number

Chemical reaction

               K₂Cr₂O₇  +  H₂O  +  S  ⇒   KOH  +  Cr₂O₃  +  SO₂    

Process

1.- Calculate the oxidation numbers following the rules.

Some rules

  H = +1       O = -2    Alkali metals = + 1    Alkali earth metals = +2    

K₂⁺¹Cr₂⁺⁶O₇⁻²  +  H₂⁺¹O⁻²  +  S⁰  ⇒   K⁺¹O⁻²H⁺¹  +  Cr₂⁺³O₃⁻²  +  S⁺⁴O₂⁻²    

Elements that changed their oxidation numbers

                       Cr₂⁺⁶   ---------------- Cr₂⁺³

                       S⁰       --------------- S⁺⁴

                     

 

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