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Ilia_Sergeevich [38]
3 years ago
10

Write a balanced chemical equation for the reaction. feso4 and pb(no3)2 express your answer as a chemical equation. identify all

of the phases in your answer.
Chemistry
2 answers:
guapka [62]3 years ago
6 0

<u>Answer:</u> The balanced chemical equation is written above.

<u>Explanation:</u>

Balanced chemical equation is defined as the equation in which total number of individual atoms on the reactant side will be equal to the total number of individual atoms on the product side. These equations follow law of conservation of mass.

When ferrous sulfate reacts with lead (II) nitrate, leads to the production of iron (II) nitrate and lead (II) sulfate.

The chemical equation for the above reaction follows:

FeSO_4(aq.)+Pb(NO_3)_2(aq.)\rightarrow Fe(NO_3)_2(aq.)+PbSO_4(s)

By Stoichiometry of the reaction:

1 mole of aqueous solution of ferrous sulfate reacts with 1 mole of aqueous solution of lead (II) nitrate to produce 1 mole of aqueous solution of iron (II) nitrate and 1 mole of solid lead (II) sulfate.

Hence, the balanced chemical equation is written above.

gulaghasi [49]3 years ago
3 0
FeSO₄(aq) + Pb(NO₃)₂(aq) → Fe(NO₃)₂(aq) + PbSO₄(s)
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The heats of combustion of ethane (C2H6) and butane (C4H10) are 52 kJ/g and 49 kJ/g, respectively. We need to produce 1.000 x 10
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Answer :

(1) The number of grams needed of each fuel (C_2H_6)\text{ and }(C_4H_{10}) are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel (C_2H_6)\text{ and }(C_4H_{10}) are 0.641 moles and 0.352 moles respectively.

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C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O

C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

(4) The number of moles of CO_2 produced by burning each fuel is 1.28 mole and 1.41 mole respectively.

The fuel that emitting least amount of CO_2 is C_2H_6

Explanation :

<u>Part 1 :</u>

First we have to calculate the number of grams needed of each fuel (C_2H_6)\text{ and }(C_4H_{10}).

As, 52 kJ energy required amount of C_2H_6 = 1 g

So, 1000 kJ energy required amount of C_2H_6 = \frac{1000}{52}=19.23g

and,

As, 49 kJ energy required amount of C_4H_{10} = 1 g

So, 1000 kJ energy required amount of C_4H_{10} = \frac{1000}{49}=20.41g

<u>Part 2 :</u>

Now we have to calculate the number of moles of each fuel (C_2H_6)\text{ and }(C_4H_{10}).

Molar mass of C_2H_6 = 30 g/mole

Molar mass of C_4H_{10} = 58 g/mole

\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles

and,

\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles

<u>Part 3 :</u>

Now we have to write down the balanced chemical equation for the combustion of the fuels.

The balanced chemical reaction for combustion of C_2H_6 is:

C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O

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The balanced chemical reaction for combustion of C_4H_{10} is:

C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

<u>Part 4 :</u>

Now we have to calculate the number of moles of CO_2 produced by burning each fuel to produce 1000 kJ.

C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O

From this we conclude that,

As, 1 mole of C_2H_6 react to produce 2 moles of CO_2

As, 0.641 mole of C_2H_6 react to produce 0.641\times 2=1.28 moles of CO_2

and,

C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

From this we conclude that,

As, 1 mole of C_4H_{10} react to produce 4 moles of CO_2

As, 0.352 mole of C_4H_{10} react to produce 0.352\times 4=1.41 moles of CO_2

So, the fuel that emitting least amount of CO_2 is C_2H_6

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