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3 years ago

If 5.32 mols N2 and 15.8 mols H2 react together, what mass NH3 can be

Chemistry

1 answer:

hodyreva [135]3 years ago

3 0

Answer:

2.87 gram

N2 is the limiting agent

Explanation:

We will find out if there is sufficient N2 and h2 to produce NH3

a) For 2.36 grams of N2

Molar mass of N2 = 28.02

Number of moles of N2 in 2.36 grams = 2.36/28.02

Mass of NH3 = 17.034 g

Now NH3 produced form 2.36 grams of N2 =

2.36/28.02 * 2 * 17.034 = 2.87 g NH3

b) For 1.52 g of H2

NH3 produced = 1.52/2.016 * (2/3) * 17.034 = 8.56

N2 Is not enough to produce 2.87 g of NH3 and also H2 is not enough to make 8.56 g of NH3.

N2 is the limiting agent as it has smaller product mass

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The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook:

german

Answer:

For 1: The largest positive cell potential is of cell having 1st and 4th half reactions.

For 2: The standard electrode potential of the cell is 1.539 V

For 3: The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

Explanation:

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

We are given:

$Ag^++(aq.)+e^-\rightarrow Ag(s);E^o_{Ag^+/Ag}=0.799V\\\\Cu^{2+}+(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.337V\\\\Ni^{2+}(aq.)+2e^-\rightarrow Ni(s);E^o_{Ni^{2+}/Ni}=-0.28V\\\\Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o_{Cr^{3+}/Cr}=-0.74V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Cell having 1st and 2nd half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Copper will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-0.337=0.462V$

Cell having 1st and 3rd half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.28)=1.079V$

Cell having 1st and 4th half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.74)=1.539V$

Cell having 2nd and 3rd half reactions:

Copper has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.337-(-0.28)=0.617V$

Cell having 3rd and 4th half reactions:

Nickel has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=-0.28-(-0.74)=0.46V$