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Olin [163]
3 years ago
9

1-calculate the mass of 4 moles of Na

Chemistry
1 answer:
noname [10]3 years ago
6 0

1. 92 g

2. 5.4 g

3. 0.217 moles

<h3>Further explanation  </h3>

The mole is the number of particles(molecules, atoms, ions) contained in a substance  

1 mol = 6.02.10²³ particles

Can be formulated

N=n x No

N = number of particles

n = mol

No = Avogadro's = 6.02.10²³

Moles can also be determined from the amount of substance mass and its molar mass

\tt n=\dfrac{m}{MW}

1. the mass of 4 moles of Na(MW= 23 g/mol)

\tt mass=mol\times MW\\\\mass=4\times 23\\\\mass=92~g

2.  the mass of 0.3 moles of water(MW=18 g/mol)

\tt mass=mol\times MW\\\\mass=0.3\times 18\\\\mass=5.4~g

3. the number of moles found in 10g of ethanol C2H5OH(MW=46 g/mol) :

\tt mol=\dfrac{10}{46}\\\\mol=0.217

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Problem PageQuestionSteam reforming of methane ( ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas,
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The question is incomplete. Her eis the complete question.

Steam reforming methane  (CH4) produces "synthesis gas", a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 125L tank with 20 mol of methane gas and 10 mol of water vapor at 38°C. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of gas hydrogen to be 18 mol. Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.

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CH_{4} + H_{2}O ⇒ CO_{} + 3H_{2}

To calculate the concentration equilibrium constant, first calculate the molarity (\frac{mol}{L}) of each molecule of the reaction.

At 38°C: At the initial temperature, there no products yet

<u>Molarity of CH4</u>:

CH4 = \frac{20}{125} = 0.16M

<u>Molarity of H20</u>:

H2O = \frac{10}{125} = 0.08M

At final temperature:

<u>Molarity of H2</u>:

H2 = \frac{18}{125} = 0.144M

According to the chemical reaction, the combination of 1 mol of each reagents produces 1 mol of CO and 3 mols of H2, so, for the products, the ratio is 1:3.

<u>Molarity of CO</u>:

CO = \frac{0.144}{3} = 0.048M

For the reagents, the proportion is 1:1, but they had an initial concentration, so, when in equilibrium, the concentration will be:

<u>Molarity of CH4</u>:

CH4 = 0.16 - 0.048 = 0.112M

<u>Molarity of H2O</u>:

H20 = 0.08 - 0.048 = 0.032M

The equilibrium constant is given by:

K_{c} = \frac{[CO][H_{2}]^{3} }{[CH_{4}][H_{2}O ] }

K_{c} = \frac{0.048.0.144^{3} }{0.112.0.032}

K_{c} = 2.10^{-2}

The concentration equilibrium constant for the process is K_{c} = 2.10^{-2}.

4 0
3 years ago
How many moles of iron(III) hydroxide precipitate will form when 2.7 moles of aqueous sodium hydroxide reacts completely with ex
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Answer:

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Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

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Now, we can determine the moles of iron (III) hydroxide formed from the reaction as follow:

From the balanced equation above,

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Therefore, 2.7 moles of NaOH will react to produce = 2.7/3 = 0.9 mole of Fe(OH)3.

Therefore, 0.9 mole of Fe(OH)3 is produced from the reaction.

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