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dlinn [17]
3 years ago
7

What is the total mass of 2.0 moles of diatomic hydrogen?

Chemistry
2 answers:
sladkih [1.3K]3 years ago
7 0

diatomic hydrogen is written as H2                                                                                (2.02 grams H2) <------- if each hydrogen atom is 1.01 grams, then two hydrogen atoms are 2.02 grams                                                                                                                  2.0 moles H2 X 2.02 grams H2                                                                                             ------------- (divide to cancel moles)   = 4.04 grams/mole H2 ÷ one mole = 4.04 grams H2

                                                          

Brut [27]3 years ago
7 0

Explanation:

A diatomic molecule is defined as the molecule which contains two atoms.

For example, a H_{2} molecule has two atoms of hydrogen.

Molar mass of a H_{2} molecule is 2.0 g/mol. Therefore, calculate the mass of 2.0 moles of diatomic hydrogen as follows.

        No. of moles = \frac{mass}{\text{molar mass}}

                2.0 mol = \frac{mass}{2.0 g/mol}

                mass = 1 g

therefore, we can conclude that total mass of 2.0 moles of diatomic hydrogen is 1 g.

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A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s
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<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %

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

  • <u>For NaBr:</u>

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol

The chemical equation for the reaction of 1-butanol and NaBr is:

\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = \frac{1}{1}\times 0.108=0.108 moles of 1-bromobutane

  • Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g

  • To calculate the percentage yield of 1-bromobutane, we use the equation:

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

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%

Hence, the percent yield of the 1-bromobutane is 48.65 %

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2 years ago
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