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adell [148]
3 years ago
12

C. If 150 grams of oxygen are produced, how many moles of sodium chlorate decomposed

Chemistry
1 answer:
Black_prince [1.1K]3 years ago
7 0

Answer:

3.13 moles of sodium chlorate are decomposed.

Explanation:

Given data:

Mass of oxygen produced = 150 g

Number of moles of sodium chlorate decomposed = ?

Solution:

Chemical equation:

2NaClO₃   →   2NaCl + 3O₂

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 150 g/ 32 g/mol

Number of moles = 4.7 mol

Now we will compare the moles of oxygen with sodium chlorate.

                     O₂           :           NaClO₃

                      3             :              2

                    4.7            :          2/3×4.7 = 3.13 mol

3.13 moles of sodium chlorate are decomposed.

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2. How many moles are in 2.8 Liters of CO2 gas?
andrew-mc [135]

Answer:

0.125 moles

Explanation:

2.8 litres is equivalent to 2.8dm³

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1 mole = 22.4 dm³

x mole = 2.8 dm³

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22.4x = 2.8

Divide both sides by 22.4

x = 2.8/22.4

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4 0
2 years ago
. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are
olga nikolaevna [1]

<u>Answer:</u> The standard free energy change of formation of S^{2-}(aq.) is 92.094 kJ/mol

<u>Explanation:</u>

We are given:

K_{sp}\text{ of }Ag_2S=8\times 10^{-51}

Relation between standard Gibbs free energy and equilibrium constant follows:

\Delta G^o=-RT\ln K

where,

\Delta G^o = standard Gibbs free energy = ?

R = Gas constant = 8.314J/K mol

T = temperature = 25^oC=[273+25]K=298K

K = equilibrium constant or solubility product = 8\times 10^{-51}

Putting values in above equation, we get:

\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ

For the given chemical equation:

Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)

The equation used to calculate Gibbs free change is of a reaction is:  

\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]

The equation for the Gibbs free energy change of the above reaction is:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]

We are given:

\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ

Putting values in above equation, we get:

285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol

Hence, the standard free energy change of formation of S^{2-}(aq.) is 92.094 kJ/mol

8 0
3 years ago
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