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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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I need this answer quick please show work
Ainat [17]

Answer:

The answer to your question is 25.2 g of acetic acid.

Explanation:

Data

[Acetic acid] = 0.839 M

Volume = 0.5 L

Molecular weight = 60.05 g/mol

Process

1.- Calculate the number of moles of acetic acid

    Molarity = moles / volume

-Solve for moles

    moles = Molarity x volume

-Substitution

    moles = (0.839)(0.5)

-Result

    moles = 0.4195

2.- Calculate the mass of acetic acid using proportions and cross multiplications

                   60.05 g ----------------------- 1 mol

                        x        ----------------------- 0.4195 moles

                        x = (0.4195 x 60.05) / 1

                        x = 25.19 g

3.- Conclusion

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