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kenny6666 [7]
3 years ago
8

A piece of solid carbon dioxide, with a mass of 7.8g,is placed in a 4.0-L otherwise empty container at 278C. What is the pressur

e in the container after all the carbon dioxide vapor- izes? If 7.8 g solid carbon dioxide were placed in the same container but it already contained air at 740 torr, what would be the partial pressure of carbon dioxide and the total pressure in the container after the carbon dioxide vaporizes?
Chemistry
1 answer:
sukhopar [10]3 years ago
7 0

Answer:

First question: P = 2.00 atm (1520 torr)

Second question: Partial pressure of carbon dioxide: 2.00 atm (1520 torr); total pressure: 2.97 atm (2260 torr).

Explanation:

The molar mass of caron dioxide, CO₂, is:

12 g/mol of C + 2*16 g/mol of O = 44 g/mol

The number of moles is the mass divided by the molar mass:

n = 7.8/44 = 0.1773 mol

By the ideal gas law:

PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.082 atm*L/mol*K), and T is the temperature (278ºC + 273 = 551 K).

So, the pressure will be:

P*4 = 0.1773*0.082*551

P = 2.00 atm = 1520 torr

If there was air in the container, the partial pressure of the carbon dioxide will still the same, 2 atm (1520 torr), because the partial pressure is the pressure the substance would have if it eas alone in the container.

The total pressure is the sum of the partial pressures, so

Pt = 740 + 1520 = 2260 torr = 2.97 atm

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<span>Answer 1: 10.03 g of siver metal can be formed.</span>
Answer 2: 3.11 g of Co are left over.

Work:

1) Unbalanced chemical equation (given):

<span>Co + AgNO3 → Co(NO3)2 + Ag

2) Balanced chemical equation
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<span>Co + 2AgNO3 → Co(NO3)2 + 2Ag

3) mole ratios

1 mol Co : 2 mole AgNO3 : 1 mol Co(NO3)2 : 2 mol Ag

4) Convert the masses in grams of the reactants into number of moles

4.1) 5.85 grams of Co

# moles = mass in grams / atomic mass

atomic mass of Co = 58.933 g/mol

# moles Co = 5.85 g / 58.933 g/mol = 0.0993 mol

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# moles Ag(NO3) = mass in grams / molar mass

molar mass AgNO3 = 169.87 g/mol

# moles Ag(NO3) = 15.8 g / 169.87 g/mol = 0.0930 mol

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Given the mole ratio 1 mol Co : 2 mol Ag(NO3) you can conclude that there is not enough Ag(NO3) to make all the Co react.

That means that Ag(NO3) is the limiting reactant, which means that it will be consumed completely, whilce Co is the excess reactant.

6) Product formed.

Use this proportion:

2 mol Ag(NO3)           0.0930mol Ag(NO3)    
--------------------- =      ---------------------------
    2 mol Ag                              x

=> x = 0.0930 mol

Convert 0.0930 mol Ag to grams:

mass Ag = # moles * atomic mass = 0.0930 mol * 107.868 g/mol = 10.03 g

Answer 1: 10.03 g of siver metal can be formed.

6) Excess reactant left over

    1 mol Co                             x
----------------------- =  ----------------------------
2 mole Ag(NO3)       0.0930 mol Ag(NO3)

=> x = 0.0930 / 2 mol Co = 0.0465 mol Co reacted

Excess = 0.0993 mol - 0.0465 mol = 0.0528 mol

Convert to grams:

0.0528 mol * 58.933 g/mol = 3.11 g

Answer 2: 3.11 g of Co are left over.
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Explanation:

Dilution is the procedure to prepare a less concentrated solution from a more concentrated one, and simply consists of adding more solvent. So, in a dilution the amount of solute does not vary, but the volume of the solvent varies.

In summary, a dilution is a lower concentration solution than the original.

The way to do the calculations in a dilution is through the expression:

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