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Arada [10]
3 years ago
8

A mixture contains methane (CH4), ethane (C2H6), and propane (C3H8). The mixture contains 25.9% methane, 31.4% ethane, and the b

alance propane by mass. Converting the mass percents to mole fractions for all three components gives:
Chemistry
1 answer:
Eddi Din [679]3 years ago
4 0

Answer:

Mol fraction methane = 0.4453

Mol fraction ethane = 0.2878

Mol fraction propane = 0.2669

Explanation:

Step 1: Data given

Suppose the mass of the mixture = 100 grams

The mixture contains:

25.9 % methane = 25.9 grams

31.4 % ethane = 31.4 grams

propane = 100 - 25.9 - 31.4 = 42.7 grams

Molar mass of methane = 16.04 g/mol

Molar mass of ethane = 30.07 g/mol

Molar mass of propane = 44.1 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles methane = 25.9 grams / 16.04 g/mol

Moles methane = 1.615 moles

Moles ethane = 31.4 grams / 30.07 g/mol

Moles ethane = 1.044 moles

Moles propane = 42.7 grams / 44.1 g/mol

Moles propane = 0.968 moles

Step 3: Calculate total moles

Total moles = moles methane + moles ethane + moles propane

Total moles = 1.615 + 1.044 + 0.968

Total moles = 3.627 moles

Step 4: Calculate mo lfraction

Mol fraction = mol component / total moles

Mol fraction methane = 1.615 moles/ 3.627 moles

Mol fraction methane = 0.4453

Mol fraction ethane = 1.044 moles / 3.627 moles

Mol fraction ethane = 0.2878

Mol fraction propane = 0.968 moles / 3.627 moles

Mol fraction propane = 0.2669

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During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of
almond37 [142]

Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate \rm CaCO_3?

Refer to a modern periodic table for relative atomic mass data:

  • Ca: 40.078;
  • C: 12.011;
  • O: 15.999.

Formula mass of \rm CaCO_3:

M(\mathrm{CaCO_3})  = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}.

\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol.

How many moles of \rm CaCl_2 will be produced?

The coefficient in front of \rm CaCO_3 in the chemical equation is the same as that in front of \rm CaCl_2. That is:

\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1.

\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol.

What's the theoretical yield of calcium chloride? In other words, what's the mass of \rm 0.949184\;mol of \rm CaCl_2?

Again, refer to a periodic table for relative atomic data:

  • Ca: 40.078;
  • Cl: 35.45.

M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}.

\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}.

What's the actual yield of calcium chloride?

\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%.

\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}.

8 0
3 years ago
True or false? different reactions require different catalysts.
lapo4ka [179]

True, different reactions require different catalysts. Hence, option 1 is correct.

<h3>What are catalysts?</h3>

A catalyst is a substance that speeds up a chemical reaction or lowers the temperature or pressure needed to start one, without itself being consumed during the reaction.

Though a catalyst is supposed to remain unaltered at the end of the reaction, it does take part in the reaction by providing active centres for the reaction to take place.

By helping to form a suitable activated complex in the course of the reaction, the catalyst increases the rate of formation of the product, as well as its yield.

Therefore, a catalyst has to be reactant-specific to form the favourable activated complex or intermediate.

Hence, option 1 is correct.

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schepotkina [342]

Answer:

Look at the picture.

Explanation:

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7 0
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To separate a mixture of iron filings and salt, the most efficient method would be
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Answer:

Explanation:

Of course you could do the separation chemically. Dissolve the salt up in water, pass thru a filter, wash the iron filings with ethanol, which would encourage the salt to precipitate from solution.

I do hope I helped you! :)

4 0
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Answer:

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