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pochemuha
3 years ago
7

How many grams of oxygen are required to produce

Chemistry
1 answer:
Likurg_2 [28]3 years ago
8 0

Answer:

8 GRAMS OF OXYGEN WILL BE REQUIRED TO PRODUCE 9 GRAMS OF WATER MOLECULES.

Explanation:

The equation for the reaction is;

2 H + O2 ----> 2 H2O          

(H = 1, O = 16)

From the reaction, it can be observed that 1 mole of O2 reacts to form 2 moles of H2O

At STP, using the molar masses of the elememts and compound in the reaction, 32 g of O2 will reacts to form 36 g (18 * 2 g) of H2O

32 g of O2 = 36 g OF H2O

if 9 grams of H2O is produced, the mass of oxygen required to produce it will be:

( 32 * 9 / 36 ) g of O2

= (288 / 36) g

= 8 g of O2

So, 8 grams of Oxygen will be required to produce 9 grams of water.

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A chemist fills a reaction vessel with 7.92 atm nitrogen (N2) gas, 2.02 atm hydrogen (H2) gas, and 2.11 atm ammonia (NH3) gas at
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<u>Answer:</u> The Gibbs free energy of the given reaction is -40 kJ

<u>Explanation:</u>

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

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(NH_3(g))})]-[(1\times \Delta G^o_f_{(N_2(g))})+(3\times \Delta G^o_f_{(H_2(g))})]

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\Delta G^o_f_{(NH_3(l))}=-16.45kJ/mol\\\Delta G^o_f_{(H_2(g))}=0kJ/mol\\\Delta G^o_f_{(N_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (-16.45))]-[(1\times (0))+(3\times (0))]\\\\\Delta G^o_{rxn}=-32.9kJ/mol

The equation used to Gibbs free energy of the reaction follows:

\Delta G=\Delta G^o+RT\ln Q_{eq}

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\Delta G = free energy of the reaction

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

Q_{eq} = Ratio of concentration of products and reactants at any time = \frac{(p_{NH_3})^2}{(p_{H_2})^3\times p_{N_2}}

p_{NH_3}=2.11atm

p_{N_2}=7.92atm

p_{H_2}=2.02atm

Putting values in above equation, we get:

\Delta G=-32900J/mol+(8.314J/K.mol\times 298K\times \ln (\frac{(2.11)^2}{(2.02)^3\times 7.92}))\\\\\Delta G=-39553.04J/mol=--39.55kJ=-40kJ

Hence, the Gibbs free energy of the given reaction is -40 kJ

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