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Vlada [557]
3 years ago
5

Polychlorinated biphenyls (PCBs), which were formerly used in the manufacture of electrical transformers, are environmental and

health hazards. They break down very slowly in the environment. The decomposition of PCBs can be represented by the equation:
Chemistry
2 answers:
jolli1 [7]3 years ago
5 0

Answer:

The decomposition of PCBs is given as:

2C_{12}H_{4}Cl_{16}+23O_2+2H_2O \rightarrow 24CO_2 + 12HCl

Explanation:

Polychlorinated biphenyls are synthetic chemicals including two benzene rings with chlorine atom attached to the ring. Yellow in color with no taste and odor.

The decomposition of PCBs is given as:

2C_{12}H_{4}Cl_{16}+23O_2+2H_2O \rightarrow 24CO_2 + 12HCl

According to reaction, 2 moles of PCBS reacts with 23 moles of oxygen gas and 2 moles of water to gives 24 moles of carbon dioxide and 12 moles of hydrochloric acid.

lubasha [3.4K]3 years ago
4 0
<span> C12H4Cl6 + 23 O2 + 2 H2O → 24 CO2 + 12 HCl </span>

<span>(a) </span>
<span>(13.5 mol O2) x (2 mol H2O / 23 mol O2) = 1.17 mol H2O </span>

<span>(b) </span>
<span>(17.6 mol H2O) x (12 mol HCl / 2 mol H2O) x (36.4611 g HCl/mol) = 3850 g HCl </span>

<span>(c) </span>
<span>(90.4 g HCl) / (36.4611 g HCl/mol) x (24 mol CO2 / 12 mol HCl) = 4.96 mol CO2 </span>

<span>(d) </span>
<span>(106.01 g CO2) / (44.00964 g CO2/mol) x (2 mol C12H4Cl6 / 24 mol CO2) x (360.8782 g C12H4Cl6/mol) = </span>
<span>72.4 g C12H4Cl6 </span>

<span>(e) </span>
<span>(4.2 kg C12H4Cl6) / (360.8782 g C12H4Cl6/mol) x (12 mol HCl / 2 mol C12H4Cl6) x (36.4611 g HCl/mol) = </span>
<span>2.5 kg HCl</span>
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ss7ja [257]

Answer:

1.89 of Sodium Carbonate

3.94 g of Silver Carbonate

2.43 g of Sodium Nitrate

Zero grams of Silver Nitrate

Explanation:

We have to start with the reaction:

AgNO_3~+~Na_2CO_3~->~Ag_2CO_3~+~NaNO_3

Now, we can balance the reaction:

2AgNO_3~+~Na_2CO_3~->~Ag_2CO_3~+~2NaNO_3

Now, we have to calculate the limiting reagent and we have to follow a few steps:

1) Convert to moles (using the molar mass of each compound)

2) Divide by the coefficient of each reactive (given by the balanced reaction)

<u>Convert to moles</u>

<u />

3.40~g~Na_2CO_3\frac{105.98~g~Na_2CO_3}{1~mol~Na_2CO_3}=0.032~mol~Na_2CO_3

4.86~g~AgNO_3\frac{169.8~g~AgNO_3}{1~mol~AgNO_3}=0.0286~mol~AgNO_3

<u>Divide by the coefficient</u>

<u></u>

\frac{0.032~mol~Na_2CO_3}{1}=0.032

<u />

\frac{0.0286~mol~AgNO_3}{2}=0.0143

The smallest value is for AgNO_3 , therefore the 4.86 g of AgNO_3 .

Now we can calculate the amount of compounds produced is we follow a few steps:

1) Use the molar ratio

2) Convert to moles (using the molar mass of each compound)

<u>Amount of Silver Carbonate</u>

<u />

0.0286~mol~AgNO_3\frac{1~mol~AgCO_3}{2~mol~AgNO_3}\frac{275.74~g~AgCO_3}{1~mol~AgCO_3}=3.94~g~AgCO_3

<u>Amount of Sodium Nitrate</u>

<u />

0.0286~mol~AgNO_3\frac{2~mol~NaNO_3}{2~mol~AgNO_3}\frac{84.99~g~NaNO_3}{1~mol~NaNO_3}=2.43~g~NaNO_3

<u>Amount of Sodium Carbonate (Excess reactive)</u>

<u />

0.0286~mol~AgNO_3\frac{1~mol~NaCO_3}{2~mol~AgNO_3}\frac{105.98~g~NaCO_3}{1~mol~NaCO_3}=1.51~g~NaCO_3

3.4~g~NaCO_3-1.51~g~NaCO_3=1.89~g~NaCO_3

<u>Amount of Silver Nitrate</u>

<u />

All the silver nitrate would be consumed in the reaction

I hope it helps!

<u />

<u />

<u />

7 0
4 years ago
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