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Galina-37 [17]
3 years ago
8

A 115.0-g sample of oxygen was produced by heating 400.0 g of potassium chlorate.2KClO3 Right arrow. 2KCI + 3O2What is the perce

nt yield of oxygen in this chemical reaction?Use Percent yield equals StartFraction actual yield over theoretical yield EndFraction times 100..
Chemistry
1 answer:
STALIN [3.7K]3 years ago
5 0

Answer:

73.4% is the percent yield

Explanation:

2KClO₃ →  2KCl  + 3O₂

This is a decomposition reaction, where 2 moles of potassium chlorate decompose to 2 moles of potassium chloride and 3 moles of oxygen.

We determine the moles of salt: 400 g . 1. mol /122.5g= 3.26 moles of KClO₃

In the theoretical yield of the reaction we say:

2 moles of potassium chlorate can produce 3 moles of oxygen

Therefore, 3.26 moles of salt, may produce (3.26 . 3) /2 = 4.89 moles of O₂

The mass of produced oxygen is: 4.89 mol . 32 g /1mol = 156.6g

But, we have produced 115 g. Let's determine the percent yield of reaction

Percent yield = (Produced yield/Theoretical yield) . 100

(115g / 156.6g) . 100 = 73.4 %

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A) Compute the repeat unit molecular weight of polystyrene. B) Compute the number-average molecular weight for a polystyrene for
Andrej [43]

Answer:

Explanation:

In Polystrene, the molecular formula for the repeat unit = C_8H_8;

and the atomic weights of Carbon C = 12.01 g/mol

For Hydrogen, it is 1.01 g/mol

Hence, the repeat unit molecular weight is:

m = 8 (12.01 g/mol)+8(1.01 g/mol)

m = 96.08 g/mol + 8.08 g/mol

m = 104.16 g/mol

The degree of polymerization = no-average molecular weight/repeat unit molecular weight.

Mathematically;

DP = \dfrac{\overline M_n}{m}

\overline M_n= DP \times m

\overline M_n= 25000 \times 104.16 \ g/mol

\overline M_n= 2604000  \ g/mol

7 0
3 years ago
Given similar concentrations, the stronger acid corresponds to the lower pH. Comment on the relative strengths of the acids H3PO
andreyandreev [35.5K]

Answer:

H3PO4 is stronger than H2PO4- because

H3PO4 dissociation constant is 6.9×10^-3

H2PO4^- dissociation constant is 6.2×10^-8

7 0
2 years ago
In which group of the Periodic Table do most of the elements exhibit both positive and negative oxidation states?
Shtirlitz [24]
C i think but you should pick it anyway
4 0
3 years ago
. A container with a volume of 0.50 L contains air at a pressure of 1.0 atm. If the volume is reduced to 0.10 L at constant temp
marin [14]

Answer:

The answer to your question is Pressure = 5 atm

Explanation:

Data

Volume 1 = V1 = 0.5 l

Pressure 1 = P1 = 1 atm

Volume 2 = V2 = 0.1 l

Pressure 2 = P2 = x

Formula

To solve this problem use the Boyle's equation

                      V1P1 = V2P2

Solve for P2

                     P2 = V1P1/V2

Substitution

                     P2 = (0.5 x 1) / 0.1

Simplification

                     P2 = 0.5/0.1

Result

                     P2 = 5 atm          

3 0
3 years ago
For the following reaction, 4.21 grams of hydrogen gas are allowed to react with 10.6 grams of ethylene (C2H4) . hydrogen(g) + e
sergiy2304 [10]

Answer:a)  11.34 g of ethane (C_2H_6) can be formed

b) C_2H_4 is the limiting reagent

c) 3.44 g of the excess reagent remains after the reaction is complete

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}    

1. \text{Moles of} H_2=\frac{4.21}{2}=2.10moles

2. \text{Moles of} C_2H_4=\frac{10.6}{28}=0.378moles

H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)

According to stoichiometry :

1 mole of C_2H_4 require 1 mole of H_2

Thus 0.378 moles of C_2H_4 will require=\frac{1}{1}\times 0.378=0.378moles  of H_2

Thus C_2H_4 is the limiting reagent as it limits the formation of product and H_2 is the excess reagent.

moles of H_2 left = (2.10-0.378) = 1.72 moles

mass of H_2 left=moles\times {\text {Molar mass}}=1.72moles\times 2g/mol=3.44g

According to stoichiometry :

As 1 mole of C_2H_4 give = 1 mole of C_2H_6

Thus 0.378 moles of C_2H_4 give =\frac{1}{1}\times 0.378=0.378moles  of C_2H_6

Mass of C_2H_6=moles\times {\text {Molar mass}}=0.378moles\times 30g/mol=11.34g

Thus 11.34 g of ethane is formed.

4 0
4 years ago
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