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2 years ago

Please help balance _Ch7H16+_O2=_CO2+_H2O

Chemistry

1 answer:

iren2701 [21]2 years ago

8 0

Answer:

C7H16 + 11 O2 → 7 CO2 + 8 H2O

This is an oxidation-reduction (redox) reaction:

7 C-16/7 - 44 e- → 7 CIV (oxidation)

22 O0 + 44 e- → 22 O-II (reduction)

C7H16 is a reducing agent, O2 is an oxidizing agent.

Reactants:

C7H16

Names: Dioxygen source: wikidata, accessed: 2019-09-07, Oxygen source: ICSC, accessed: 2019-09-04source: wikidata, accessed: 2019-09-07, Oxygen (liquefied) source: ICSC, accessed: 2019-09-04

Appearance: Odourless compressed gas source: ICSC, accessed: 2019-09-04; Liquefied gas. colourless-to-blue extremely cold liquid source: ICSC, accessed: 2019-09-04

Products:

CO2

Names: Carbon dioxide source: wikipedia, accessed: 2019-09-27source: wikidata, accessed: 2019-09-02source: ICSC, accessed: 2019-09-04source: NIOSH NPG, accessed: 2019-09-02, {{plainlist| source: wikipedia, accessed: 2019-09-27, CO2 source: wikidata, accessed: 2019-09-02

Appearance: Colorless gas source: wikipedia, accessed: 2019-09-27; Odourless colourless compressed liquefied gas source: ICSC, accessed: 2019-09-04; Colorless, odorless gas. [Note: Shipped as a liquefied compressed gas. Solid form is utilized as dry ice.] source: NIOSH NPG, accessed: 2019-09-02

H2O – Water, oxidane source: wikipedia, accessed: 2019-09-27

Other names: Water (H2O) source: wikipedia, accessed: 2019-09-27, Hydrogen hydroxide (HH or HOH) source: wikipedia, accessed: 2019-09-27, Hydrogen oxide source: wikipedia, accessed: 2019-09-27

Appearance: White crystalline solid, almost colorless liquid with a hint of blue, colorless gas source: wikipedia, accessed: 2019-09-27

Search by reactants (C7H16, O2)

1 O2 + C7H16 → H2O + CO2

2 O2 + C7H16 → H2O + CO

3 O2 + C7H16 → H2O + CO2 + CO

4 O2 + C7H16 → H2 + CO2

Search by products (CO2, H2O)

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1 HCl + CaCO3 → H2O + CO2 + CaCl2

2 HCl + NaHCO3 → H2O + CO2 + NaCl

3 O2 + CH4 → H2O + CO2

4 NaHCO3 → H2O + CO2 + Na2CO3

5 H2SO4 + KMnO4 + (COOH)2 → H2O + CO2 + K2SO4 + MnSO4

6 O2 + C3H8 → H2O + CO2

7 H2SO4 + K2CO3 → H2O + CO2 + K2SO4

8 O2 + C2H6 → H2O + CO2

9 CH3COOH + NaHCO3 → H2O + CO2 + CH3COONa

10 HCl + Na2CO3 → H2O + CO2 + NaCl

12345678910...

Search by reactants (C7H16, O2) and by products (CO2, H2O)

1 O2 + C7H16 → H2O + CO2

2 O2 + C7H16 → H2O + CO2 + CO

Explanation:

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5 moles of an ideal gas (Cp = 3R, Cv 2R ) are heated from 25°C to 300°C. Calculate (a) The change in internal energy of the gas

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(a) The change in internal energy of the gas is 22.86 kJ.

(b) The change in enthalpy of the gas is 34.29 kJ.

Explanation :

(a) The formula used for change in internal energy of the gas is:

$\Delta U=nC_v\Delta T\\\\\Delta U=nC_v(T_2-T_1)$

where,

$\Delta U$ = change in internal energy = ?

n = number of moles of gas = 5 moles

$C_v$ = heat capacity at constant volume = 2R

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $25^oC=273+25=298K$

$T_2$ = final temperature = $300^oC=273+300=573K$

Now put all the given values in the above formula, we get:

$\Delta U=nC_v(T_2-T_1)$

$\Delta U=(5moles)\times (2R)\times (573-298)$

$\Delta U=(5moles)\times 2(8.314J/mole.K)\times (573-298)$

$\Delta U=22863.5J=22.86kJ$

The change in internal energy of the gas is 22.86 kJ.

(b) The formula used for change in enthalpy of the gas is:

$\Delta H=nC_p\Delta T\\\\\Delta H=nC_p(T_2-T_1)$

where,

$\Delta H$ = change in enthalpy = ?

n = number of moles of gas = 5 moles

$C_p$ = heat capacity at constant pressure = 3R

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $25^oC=273+25=298K$

$T_2$ = final temperature = $300^oC=273+300=573K$

Now put all the given values in the above formula, we get:

$\Delta H=nC_p(T_2-T_1)$

$\Delta H=(5moles)\times (3R)\times (573-298)$

$\Delta H=(5moles)\times 3(8.314J/mole.K)\times (573-298)$

$\Delta H=34295.25J=34.29kJ$

The change in enthalpy of the gas is 34.29 kJ.

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Concentration of $Cr_2O_7^{2-}$ = 0.03101 M

Concentration of $MnO_4^-$ = 0.03721 M

Explanation:

The reduction for $Cr_2O_7^{2-}$ is;

$Cr_2O_7^{2-} + 14 H ^+ _{(aq)} + 6 e^- -----> 2 Cr^{3+} _{(aq)}+7H_2O _{(l)}$

$Cu^+_{(aq)} -----> Cu^{2+} _{(aq)} + 1 e^-$

6 moles of $Cu ^+$ = 1 mole of $Cr_2O_7^{2-}$

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number of moles of $Cr_2O_7^{2-}$ reacted = 0.001245 mole

Concentration of $Cr_2O_7^{2-}$ = $\frac{number \ of moles }{Volume}$

Given that the volume = 40.15 mL = $40.15 *10^{-3}$ ; we have:

Concentration of $Cr_2O_7^{2-}$ = $\frac{0.001245}{40.15*10^{-3}}$

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The reduction for $MnO_4^-$ is;

$MnO_4^- + 8H^+ + 5 e^- -----> Mn^{2+} + 4H_2O$

$Cu^+_{(aq)} -----> Cu^{2+} _{(aq)} + 1 e^-$

5 moles of $Cu ^+$ = 1 mole of $Cr_2O_7^{2-}$

number of moles of Cu reacted = $\frac{mass \ of \ Cu \ wire }{ molecular weigh tof \ Cu wire }$

number of moles of Cu reacted = $\frac{0.4749}{63.55}$

number of moles of Cu reacted = 0.00747 mole

number of moles of $MnO_4^-$ reacted = $\frac{0.00747}{5}$

number of moles of $MnO_4^-$ reacted = 0.001494 mole

Concentration of $MnO_4^-$ = $\frac{number \ of moles }{Volume}$

Given that the volume = 40.15 mL = $40.15 *10^{-3}$ ; we have:

Concentration of $MnO_4^-$ = $\frac{0.001494 }{40.15*10^{-3}}$

Concentration of $MnO_4^-$ = 0.03721 M

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