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Phoenix [80]
3 years ago
13

Given the general chemical equation 3A + 2B → C + 5D,

Chemistry
1 answer:
Irina-Kira [14]3 years ago
4 0

Answer :

(a) The number of moles of D produced can be, 6.67 moles.

(b) The volume of D prepared can be, 24.5 L

Explanation :

The given chemical reaction is:

3A+2B\rightarrow C+5D

Part (a) :

From the balanced chemical reaction, we conclude that:

As, 3 moles of A react to give 5 moles of D

So, 4 moles of A react to give \frac{5}{3}\times 4=6.67 moles of D

Thus, the number of moles of D produced can be, 6.67 moles.

Part (b) :

As we know that 1 moles of substance occupies 22.4 L volume of gas.

As, 2\times 22.4L volume of B gives 5\times 22.4L volume of D

As, 9.8 L volume of B gives \frac{5\times 22.4L}{2\times 22.4L}\times 9.8L=24.5L volume of D

Thus, the volume of D prepared can be, 24.5 L

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Sergio [31]

The full question can be seen below:

2H_{2}O_2_{(aq)} --> 2H_{2}O_{(l)}+O_{2}_{g}

The decomposition of H_{2}O_{aq} is represented by the equation above.

A student monitored the decomposition of a 1.0 L sample of H_{2}O_2_(_{aq}_) at a constant temperature of 300K and recorded the concentration of H_{2}O_2 as function of time. The results are given in the table below:

                                    Time (s)      H_{2}O_2

                                    0                 2.7

                                    200            2.1

                                    400            1.7

                                    600            1.4

The O_2_(_{g}_) produced from the decomposition of the 1.0 L sample of H_{2}O_2_(_{aq}_) is collected in a previously evacuated 10.0 L flask at 300 K. What is the approximate pressure in the flask after 400 s?

(For estimation purpose, assume that 1.0 mole of gas in 1.0 L exerts a pressure of 24 atm at 300 K).

Answer:

1.2 atm

Explanation:

Considering all assumptions as stated above;

                       2H_{2}O_2_{(aq)} --> 2H_{2}O_{(l)}+O_{2}_{g}

Initial               2.7 mole               ---             ---

Change          -1.0                        ---             +\frac{1.0}{2}

Equilibrium     1.7 mole                ---             0.5 mole

To determine the concentration of O₂; we need to convert the moles to concentration for O₂ = \frac{0.5}{volume in the flask}

                                  = \frac{0.5 mol}{10.0 L}

                                  = 0.05 \frac{mol}{L}

Thus, based on the assumption that "1.0 mole of gas in 1.0 L exerts a pressure of 24 atm"

∴ 0.05\frac{mol}{L} will give rise to = 0.05 \frac{mol}{L} × 24

                                           = 1.2 atm

7 0
3 years ago
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