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Rus_ich [418]
3 years ago
15

The balanced equation for the reaction of aqueous Pb(ClO3)2 with aqueous NaI is Pb(ClO3)2(aq)+2NaI(aq)⟶PbI2(s)+2NaClO3(aq) What

mass of precipitate will form if 1.50 L of concentrated Pb(ClO3)2 is mixed with 0.200 L of 0.210 M NaI? Assume the reaction goes to completion. mass of precipitate:
Chemistry
1 answer:
ryzh [129]3 years ago
6 0

Answer : The mass of PbI_2 precipitate produced will be, 9.681 grams.

Explanation : Given,

Molarity of NaI = 0.210 M

Volume of solution = 0.2 L

Molar mass of PbI_2 = 461.01 g/mole

First we have to calculate the moles of NaI.

\text{Moles of }NaI=\text{Molarity of }NaI\times \text{Volume of solution}=0.210M\times 0.2L=0.042moles

Now we have to calculate the moles of PbI_2.

The balanced chemical reaction is,

Pb(ClO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaClO_3(aq)

From the balanced reaction we conclude that

As, 2 moles of NaI react to give 1 mole of PbI_2

So, 0.042 moles of NaI react to give \frac{0.042}{2}=0.021 moles of PbI_2

Now we have to calculate the mass of PbI_2.

\text{Mass of }PbI_2=\text{Moles of }PbI_2\times \text{Molar mass of }PbI_2

\text{Mass of }PbI_2=(0.021mole)\times (461.01g/mole)=9.681g

Therefore, the mass of PbI_2 precipitate produced will be, 9.681 grams.

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The solution that conducts electricity and has a pH value of 7 would most likely be a neutral solution. Water is among the best examples of a neutral solution. When the pH of a solution is considered to be lesser than 7, it is an acid, while if the pH is greater than 7, it is considered to be a basic solution.
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When the following reaction reached equilibrium at 325 K , the equilibrium constant was found to be 172. When a sample was taken
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Answer: 6.88M

Explanation:

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What is the mass, in grams, 200mL of liquid with a density of 1.4g/mL
Tcecarenko [31]

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Why do noble gases not have electronegativity values
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Did this help?

6 0
3 years ago
g Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.71 g of ethane i
Kamila [148]

Answer:

6.05g

Explanation:

The reaction is given as;

Ethane + oxygen --> Carbon dioxide + water

2C2H6 + 7O2 --> 4CO2 + 6H2O

From the reaction above;

2 mol of ethane reacts with 7 mol of oxygen.

To proceed, we have to obtain the limiting reagent,

2,71g of ethane;

Number of moles = Mass / molar mass = 2.71 / 30 = 0.0903 mol

3.8g of oxygen;

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If 0.0903 moles of ethane was used, it would require;

2 = 7

0.0903 = x

x = 0.31605 mol of oxygen needed

This means that oxygen is our limiting reagent.

From the reaction,

7 mol of oxygen yields 4 mol of carbon dioxide

0.2375 yields x?

7 = 4

0.2375 = x

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8 0
3 years ago
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