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3 years ago

How many moles of hydrogen gas are produced when 0.066 mole of sodium is completely reacted?

Chemistry

2 answers:

velikii [3]3 years ago

8 0

Answer : The number of moles of hydrogen gas is, 0.033 mole

Explanation : Given,

Moles of sodium = 0.066 mole

The balanced chemical reaction will be,

$2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq)+H_2(g)$

From the balanced chemical reaction, we conclude that 2 moles of sodium react with 2 moles of water to give 2 moles of sodium hydroxide and 1 mole of hydrogen has as a product.

As, 2 moles of sodium react to give 1 mole of hydrogen gas

So, 0.066 mole of sodium react to give $\frac{1}{2}\times 0.066=0.033$ moles of hydrogen gas

Therefore, the number of moles of hydrogen gas is, 0.033 mole

SVEN [57.7K]3 years ago

6 0

Base on my research the complete reaction of hydrogen and sodium are form by this equation 2Na + 2H2O = 2NaOH +H2 If we base from this equation it shows you already the ratio of the moles of the product and reactants taking part in the reaction

1 mole of Na r/w 1 mole of water to produce .5 mole of Hydrogen

Therefore, .066 mole of Na produces .033 moles of H2

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Alcl3 + 3Na → Al + 3NaCl

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3 years ago

Two isotopes of hydrogen fuse to form a neutron plus the larger element,A) beryllium.B) carbon.C) deuterium.D) helium.

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3 0

3 years ago

4.33 g of 3-hexanol were obtained from 5.84 g of hex-3-ene. Determine the percentage yield of 3-hexanol. a Determine the moles o

Vilka [71]

<u>Answer:</u> The amount of hex-3-ene used is 0.0711 moles and the percent yield of 3-hexanol is 59.56 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of hex-3-ene = 5.84 g

Molar mass of hex-3-ene = 82.14 g/mol

Putting values in equation 1, we get:

$\text{Moles of hex-3-ene}=\frac{5.84g}{82.14/mol}=0.0711mol$

The chemical equation for the conversion of hex-3-ene to 3-hexanol follows:

$\text{hex-3-ene}+H_2O\xrightarrow []{10\% H_2SO_4} \text{3-hexanol}$

By Stoichiometry of the reaction:

1 mole of hex-3-ene produces 1 mole of 3-hexanol

So, 0.0711 moles of hex-3-ene will produce = $\frac{1}{1}\times 0.0711=0.0711mol$ of 3-hexanol

Now, calculating the mass of 3-hexanol from equation 1, we get:

Molar mass of 3-hexanol = 102.2 g/mol

Moles of 3-hexanol = 0.0711 moles

Putting values in equation 1, we get:

$0.0711mol=\frac{\text{Mass of 3-hexanol}}{102.2g/mol}\\\\\text{Mass of 3-hexanol}=(0.0711mol\times 102.2g/mol)=7.27g$

To calculate the percentage yield of 3-hexanol, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of 3-hexanol = 4.33 g

Theoretical yield of 3-hexanol = 7.27 g

Putting values in above equation, we get:

$\%\text{ yield of 3-hexanol}=\frac{4.33g}{7.27g}\times 100\\\\\% \text{yield of 3-hexanol}=59.56\%$

Hence, the amount of hex-3-ene used is 0.0711 moles and the percent yield of 3-hexanol is 59.56 %

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