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3 years ago

If 1.320 moles of CH4 reacts completely with oxygen, how many grams of H2O can be formed

Chemistry

2 answers:

Alex17521 [72]3 years ago

8 0

Answer:

mass H2O = 47.56 g

Explanation:

balanced reation:

CH4 + 2O2 → CO2 + 2H2O

∴ moles CH4 = 1.320 mol

⇒ moles H2O = (1.320 mol CH4)×(2 mol H2O/mol CH4)

⇒ moles H2O = 2.64 mol H2O

∴ molar mass (mm) H2O = 18.015 g/mol

⇒ mass H2O = (2.64 mol H2O)×(18.015 g/mol)

⇒ mass H2O = 47.56 g

Rainbow [258]3 years ago

3 0

Answer:

47.5 g of water can be formed

Explanation:

This is the reaction:

CH₄ + 2O₂ → CO₂ + 2H₂O

Methane combustion.

In this process 1 mol of methane react with 2 moles of oxygen to produce 2 moles of water and 1 mol of carbon dioxide.

As ratio is 1:2, I will produce the double of moles of water, with the moles of methane I have.

1.320 mol .2 = 2.64 moles

Now, we can convert the moles to mass (mol . molar mass)

2.64 mol . 18g/mol = 47.5 g

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2.3 Zinc has five naturally occurring isotopes: 48.63% of 64 Zn with an atomic weight of 63.929 amu; 27.90% of 66Zn with an atom

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Answer: The average atomic mass of element Zinc is 65.40 amu.

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Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

For $_{30}^{64}\textrm{Zn}$ isotope:

Mass of $_{30}^{64}\textrm{Zn}$ isotope = 63.929 amu

Percentage abundance of $_{30}^{64}\textrm{Zn}$ isotope = 48.63 %

Fractional abundance of $_{30}^{64}\textrm{Zn}$ isotope = 0.4863

For $_{30}^{66}\textrm{Zn}$ isotope:

Mass of $_{30}^{66}\textrm{Zn}$ isotope = 65.926 amu

Percentage abundance of $_{30}^{66}\textrm{Zn}$ isotope = 27.90 %

Fractional abundance of $_{30}^{66}\textrm{Zn}$ isotope = 0.2790

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Percentage abundance of $_{30}^{67}\textrm{Zn}$ isotope = 4.10 %

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Percentage abundance of $_{30}^{68}\textrm{Zn}$ isotope = 18.75 %

Fractional abundance of $_{30}^{68}\textrm{Zn}$ isotope = 0.1875

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Mass of $_{30}^{70}\textrm{Zn}$ isotope = 69.925 amu

Percentage abundance of $_{30}^{70}\textrm{Zn}$ isotope = 0.62 %

Fractional abundance of $_{30}^{70}\textrm{Zn}$ isotope = 0.0062

Putting values in equation 1, we get:

$\text{Average atomic mass of Zinc}=[(63.929\times 0.4863)+(65.926\times 0.2790)+(66.927\times 0.0410)+(67.925\times 0.1875)+(69.925\times 0.0062)]$

$\text{Average atomic mass of Zinc}=65.40amu$

Hence, the average atomic mass of element Zinc is 65.40 amu.

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