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Alekssandra [29.7K]
3 years ago
8

-

Chemistry
1 answer:
Dovator [93]3 years ago
3 0

Answer:

5 × 10^-4 L

Explanation:

The equation of the reaction is;

2KClO3 = 2KCl + 3O2

Number of moles of KClO3 = 13.5g/122.5 g / mol = 0.11 moles

From the stoichiometry of the reaction;

2 moles of KClO3 yields 3 moles of O2

0.11 moles of KClO3 yields 0.11 × 3/2 = 0.165 moles of oxygen gas

From the ideal gas equation;

PV= nRT

P= 85.4 × 10^4 KPa

V=?

n= 0.165

R= 8.314 J K-1 mol-1

T= 40+273 = 313K

V= 0.165 ×8.134 × 313/85.4 × 10^4

V=429.4/85.4 × 10^4

V= 5 × 10^-4 L

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Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb ( NO 3 ) 2
nikklg [1K]

Answer : The volume of NH_4I solution required is, 2.93 L

The number of moles of PbI_2 formed from the reaction is, 0.662 moles.

Explanation :

First we have to calculate the initial moles of Pb(NO_3)_2.

\text{Moles of }Pb(NO_3)_2=\text{Concentration of }Pb(NO_3)_2\times \text{Volume of solution}

\text{Moles of }Pb(NO_3)_2=0.700M\times 0.945L=0.662mol

Now we have to calculate the moles of NH_4I

The balanced chemical reaction is:

Pb(NO_3)_2(aq)+2NH_4I(aq)\rightarrow PbI_2(s)+2NH_4NO_3(aq)

From the balanced chemical reaction we conclude that,

As, 1 mole of Pb(NO_3)_2 react with 2 moles of NH_4I

So, 0.662 mole of Pb(NO_3)_2 react with 0.662\times 2=1.32 moles of NH_4I

Now we have to calculate the volume of NH_4I

\text{Volume of }NH_4I=\frac{\text{Moles of }NH_4I}{\text{Concentration of }NH_4I}

\text{Volume of }NH_4I=\frac{1.32mol}{0.450mol/L}=2.93L

Now we have to calculate the moles of PbI_2

From the balanced chemical reaction we conclude that,

As, 1 mole of Pb(NO_3)_2 react to give 1 moles of PbI_2

So, 0.662 mole of Pb(NO_3)_2 react to give 0.662 moles of PbI_2

Thus, the number of moles of PbI_2 formed from the reaction is, 0.662 moles.

7 0
3 years ago
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