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natka813 [3]
3 years ago
11

For the following reaction, calculate how many moles of each product are formed when 4.05 g of water is used.

Chemistry
1 answer:
timurjin [86]3 years ago
7 0

Answer:

A. 0.225 mole of H₂

B. 0.113 mole of O₂.

Explanation:

We'll begin by calculating the number of mole in 4.05 g of water (H₂O). This can be obtained as follow:

Mass of H₂O = 4.05 g

Molar mass of H₂O = (2×1) + 16 = 2 + 16 = 18 g/mol

Mole of H₂O =?

Mole = mass /Molar mass

Mole of H₂O = 4.05 / 18

Mole of H₂O = 0.225 mole

Next, the balanced equation.

2H₂O —> 2H₂ + O₂

From the balanced equation above,

2 moles of H₂O produced 2 moles of H₂ and 1 mole of O₂.

A. Determination of the number of mole hydrogen produced.

From the balanced equation above,

2 moles of H₂O produced 2 moles of H₂.

Therefore, 0.225 mole of H₂O will also produce 0.225 mole of H₂.

B. Determination of the number of mole oxygen produced.

From the balanced equation above,

2 moles of H₂O produced 1 mole of O₂.

Therefore, 0.225 mole of H₂O will produce = (0.225 × 1)/2 = 0.113 mole of O₂.

Thus, 0.113 mole of O₂ is produced.

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A. amount of precipitation, average temperature

Explanation:

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(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g
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Answer:

Part A

 The volume of the gaseous product  is  V = 787L

Part B

The volume of the the engine’s gaseous exhaust is  V_e = 2178 \ L

Explanation:

Part A

From the question we are told that

    The temperature is  T = 350^oC = 350 +273 =623K

     The pressure is  P = 735 \ torr = \frac{735}{760} =  0.967\ atm

     The of  C_8 H_{18} = 100.0g

The chemical equation for this combustion is

               2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}

 The number of moles of  C_8 H_{18} that reacted is mathematically represented as

               n = \frac{mass \ of \  C_8H_{18}  }{Molar \  mass \ of  C_8H_{18} }

The molar mass of  C_8 H_{18} is constant value which is

                  M = 114.23 \ g/mole  

So          n = \frac{100  }{114.23} }

             n = 0.8754 \ moles

The gaseous product in the reaction is CO_2_{(g)} and water vapour

Now from the reaction

    2 moles of C_8 H_{18}  will react with 25 moles of O_2 to give (16 + 18) moles of CO_2_{(g)} and  H_2 O_{(g)}

So

    1 mole of C_8 H_{18} will  react with 12.5 moles of  O_2 to give 17 moles of CO_2_{(g)} and  H_2 O_{(g)}

This implies that

    0.8754 moles of C_8 H_{18} will react with (12.5 * 0.8754 ) moles of O_2 to give  (17 * 0.8754) of CO_2_{(g)} and  H_2 O_{(g)}

So the no of moles of gaseous product is

         N_g = 17 * 0.8754

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From the ideal gas law

       PV = N_gRT

making V the subject

        V = \frac{N_gRT}{P}

Where R is the gas constant with a value R = 0.08206 \  L\cdot atm /K \cdot mole

Substituting values

          V = \frac{14.88* 0.08206 *623}{0.967}

          V = 787L

Part B

From the reaction the number of moles of oxygen that reacted is

         N_o = 0.8754 * 12.5

         N_o = 10.94 \ moles

The volume is

      V_o  = \frac{10.94 * 0.08206 *623}{0.967}

      V_o  = 579 \ L

No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

         V_e = V_o * \frac{0.79}{0.21}

Substituting values

       V_e = 579 * \frac{0.79}{0.21}

       V_e = 2178 \ L

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