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2 years ago

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

Chemistry

1 answer:

hodyreva [135]2 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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Sever21 [200]

6.11% w/v of Cu2+ implies that 6.11 g of Cu2+ is present in 100 ml of the solution

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1 mole of CuCl2 contain 1 mole of Cu2+ ion

Hence, 0.2404 moles of Cu2+ would correspond to 0.2404 moles of CuCl2

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3 years ago

What is the percent composition by mass of hydrogen in nh4hco3

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Answer: The percent composition by mass of hydrogen in given compound is 6.33 %

Explanation:

We are given:

A chemical compound having chemical formula of $NH_4HCO_3$

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To calculate the percentage composition by mass of hydrogen in the compound, we use the equation:

$\%\text{ composition of Hydrogen}=\frac{\text{Mass of hydrogen}}{\text{Mass of compound}}\times 100$

Mass of compound = $[(1\times 14)+(5\times 1)+(1\times 12)+(3\times 16)]=79g/mol$

Mass of hydrogen = $(5\times 1)=5g/mol$

Putting values in above equation, we get:

$\%\text{ composition of Hydrogen}=\frac{5g/mol}{79g/mol}\times 100=6.33\%$

Hence, the percent composition by mass of hydrogen in given compound is 6.33 %

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3 years ago

Determine the free energy(ΔG) from the standard cell potential (Ecell0 ) for the reaction:2ClO2-(aq)+Cl2(g)→2ClO2(g)+ 2Cl-(aq)wh

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Explanation:

For the given chemical reaction:

$2ClO_2^-(aq.)+Cl_2(g)\rightarrow 2ClO_2(g)+2Cl^-(aq.)$

Half reactions for the given cell follows:

Oxidation half reaction: $ClO_2^-\rightarrow ClO_2+e^-;E^o_{ClO_2^-/ClO_2}=0.954V$ ( × 2)

Reduction half reaction: $Cl_2+2e^-\rightarrow 2Cl(g);E^o_{Cl_2/2Cl^-}=1.36V$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

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

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

Putting values in above equation, we get:

$E^o_{cell}=1.36-(0.954)=0.406V$

To calculate standard Gibbs free energy, we use the equation:

$\Delta G^o=-nFE^o_{cell}$

Where,

n = number of electrons transferred = 2

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Putting values in above equation, we get:

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Hence, the $\Delta G^o$ for the given reaction is $-7.84\times 10^4J$

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