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MA_775_DIABLO [31]
2 years ago
6

According to the following balanced reaction, how many moles of HNO3 are formed from 8.44 moles of NO2 if there is plenty of wat

er present? 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)
Chemistry
1 answer:
kotegsom [21]2 years ago
3 0

Answer:

5.63 mol.

Explanation:

  • The balanced chemical equation between NO₂ and H₂O is:

<em>3NO₂(s) + H₂O(l) → 2HNO₃(aq) + NO(g), </em>

It is clear that 3 mol of NO₂ reacts with 1 mol of H₂O to produce 2 mol of HNO₃ and 1 mol of NO.

<em>Water is present as an excess reactant and NO₂ is limiting reactant.</em>

<em></em>

  • To find the no. of moles of HNO₃ produced:

3 mol of NO₂ produces → 2 mol of HNO₃, from stichiometry.

8.44 mol of NO₂ produces → ??? mol of HNO₃.

∴ The no. of moles of HNO₃ are formed = (8.44 mol)(2 mol)/(3 mol) = 5.63 mol.

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Write chemical equations and corresponding equilibrium expressions for each of the two ionization steps of carbonic acid. Part A
lesantik [10]

<u>Answer:</u> The chemical equations and equilibrium constant expression for each ionization steps is written below.

<u>Explanation:</u>

The chemical formula of carbonic acid is H_2CO_3. It is a diprotic weak acid which means that it will release two hydrogen ions when dissolved in water

The chemical equation for the first dissociation of carbonic acid follows:

               H_2CO_3(aq.)\rightleftharpoons H^+(aq.)+HCO_3^-(aq.)

The expression of first equilibrium constant equation follows:

Ka_1=\frac{[H^+][HCO_3^{-}]}{[H_2CO_3]}

The chemical equation for the second dissociation of carbonic acid follows:

               HCO_3^-(aq.)\rightarrow H^+(aq.)+CO_3^{2-}(aq.)

The expression of second equilibrium constant equation follows:

Ka_2=\frac{[H^+][CO_3^{2-}]}{[HCO_3^-]}

Hence, the chemical equations and equilibrium constant expression for each ionization steps is written above.

6 0
2 years ago
Ethyl alcohol is produced by the fermentation of glucose, C6H12O6. C6H12O6 (s) → 2 C2H5OH (l) + 2 CO2 (g) ΔH° = – 69.1 kJ Given
Norma-Jean [14]

Answer:

-1273.3

Explanation:

Enthalpy of formation of a compound is the amount of heat absorbed or evolved when one mole of the compound is formed from other compounds.

enthalpy of formation Of CO2 = 2 X -393.5 = -787

enthalpy of formation Of C2H5OH = 2 X -277.7 = -555.4

enthalpy of formation Of C6H12O6 = 69.1 (reverse sign) + (-787 + -555.4) = - 1273.3 Joules

6 0
2 years ago
Given equilibrium partial pressures of PNO2= 0.247 atm, PNO = 0.0022atm, and PO2 = 0.0011 atm calculate the equilibrium constant
maxonik [38]
Answer 1:
Equilibrium constant (K) mathematically expressed as the ratio of the concentration of products to concentration of reactant. In case of gaseous system, partial pressure is used, instead to concentration.

In present case, following reaction is involved:

                        2NO2    ↔      2NO + O2

Here, K = \frac{[PNO]^2[O2]}{[PNO2]^2}

Given: At equilibrium, <span>PNO2= 0.247 atm, PNO = 0.0022atm, and PO2 = 0.0011 atm
</span>
Hence,  K = \frac{[0.0022]^2[0.0011]}{[0.247]^2}
                 = 8.727 X 10^-8

Thus, equilibrium constant of reaction = 8.727 X 10^-8
.......................................................................................................................

Answer 2:
Given: <span>PNO2= 0.192 atm, PNO = 0.021 atm, and PO2 = 0.037 atm.

Therefore, Reaction quotient = </span>\frac{[PNO]^2[O2]}{[PNO2]^2}
                                              = \frac{[0.021]^2[0.037]}{[0.192]^2}
                                              = 4.426 X 10^-4.

Here, Reaction quotient > Equilibrium constant.

Hence, <span>the reaction need to go to reverse direction to reattain equilibrium </span>
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Explanation:

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