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3 years ago

2C2H2 + 502 = 4CO2 + 2H20when 20000g of 2C2H2 + 5O2 = how much 4CO2?

Chemistry

1 answer:

lukranit [14]3 years ago

7 0

Answer:

6.761 × 10⁴ g

Explanation:

Step 1: Write the balanced combustion reaction

2 C₂H₂ + 5 O₂ ⇒ 4 CO₂ + 2 H₂O

Step 2: Calculate the moles corresponding to 20000 g of C₂H₂

The molar mass of C₂H₂ is 26.038 g/mol.

20000 g × 1 mol/26.038 g = 768.11 mol

Step 3: Calculate the moles of CO₂ produced from 768.11 moles of C₂H₂

The molar ratio of C₂H₂ to CO₂ is 2:4.

768.11 mol C₂H₂ × 4 mol CO₂/2 mol C₂H₂ = 1536.2 mol CO₂

Step 4: Calculate the mass corresponding to 1536.2 moles of CO₂

The molar mass of CO₂ is 44.01 g/mol.

1536.2 mol × 44.01 g/mol = 6.761 × 10⁴ g

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3 years ago

You perform a distillation to separate a mixture of propylbenzene and cyclohexane, and you obtain 2.9949 grams of cyclohexane (d

Romashka-Z-Leto [24]

Answer:

66.67%

Explanation:

From the given information:

mass of cyclohexane = 2.9949 grams

density of cyclohexane = 0.779 g/mL

Recall that:

Density = mass/volume

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Volume = mass/density

So, the volume of cyclohexane = 2.9949 g/ 0.779 g/mL

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Volume of propylbenzene = 1.6575 g/ 0.862 g/mL

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The volume % composition of cyclohexane from the mixture is:

$= (\dfrac{v_{cyclohexane}}{v_{cyclohexane}+v_{propylbenzene}})\times 100$

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$= (\dfrac{3.8445}{5.7674})\times 100$

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3 years ago

Identify two ways that temperature plays a role in chemical changes.

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Explanation:

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3 years ago

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2 years ago

A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor

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Answer : The partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of $C_2HBrClF_3$ = 15.0 g

Mass of $O_2$ = 22.6 g

Molar mass of $C_2HBrClF_3$ = 197.4 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_2HBrClF_3$ and $O_2$ .

$\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole$

and,

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole$

Now we have to calculate the mole fraction of $C_2HBrClF_3$ and $O_2$ .

$\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971$