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2 years ago

Equal masses of carbon dioxide (CO2) and oxygen (O2) are mixed. What is the mole

1 answer:

6 0

The mole fraction of oxygen in this mixture is = 0.5789 which is not listed in the options. while the mole fraction of carbon dioxide = 0.4211 ( A )

Given that :

Mass of carbon dioxide = mass of oxygen

lets assume the given mass = x

first step : determine the moles of CO₂ and O₂

moles of CO₂ = mass / molar mass of CO₂

= x / 44

moles of O₂ = mass / molar mass of O2

= x / 32

∴ mole fraction of O₂ = moles of O₂ / ( Total moles in the mixture )

= $\frac{x}{32} / ( \frac{x}{44} + \frac{x}{32} )$

= $\frac{44*32*(x)}{32 * 76 *(x)}$ = 0.5789

While the mole fraction of Carbon dioxide = 1 - 0.5789 = 0.4211

hence the mole fraction of oxygen in this mixture = 0.5789 while the mole fraction of Carbon dioxide = 0.4211

learn more here : brainly.com/question/15063496

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Lana71 [14]

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4 0

2 years ago

If 34.7 g of AgNO₃ react with 28.6 g of H₂SO₄ according to this UNBALANCED equation below, how many grams of Ag₂SO₄ could be for

Luba_88 [7]

The number of grams of Ag2SO4 that could be formed is 31.8 grams

calculation

Balanced equation is as below

2 AgNO3 (aq) + H2SO4(aq) → Ag2SO4 (s) +2 HNO3 (aq)

Find the moles of each reactant by use of mole= mass/molar mass formula

that is moles of AgNO3= 34.7 g / 169.87 g/mol= 0.204 moles

moles of H2SO4 = 28.6 g/98 g/mol =0.292 moles

use the mole ratio to determine the moles of Ag2SO4

that is;

the mole ratio of AgNo3 : Ag2SO4 is 2:1 therefore the moles of Ag2SO4= 0.204 x1/2=0.102 moles

The moles ratio of H2SO4 : Ag2SO4 is 1:1 therefore the moles of Ag2SO4 = 0.292 moles

AgNO3 is the limiting reagent therefore the moles of Ag2SO4 = 0.102 moles

<h3> finally find the mass of Ag2SO4 by use of mass=mole x molar mass formula</h3>

that is 0.102 moles x 311.8 g/mol= 31.8 grams

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3 years ago

For the complete combustion of 1.000 mole of methane gas at 298 K and 1 atm pressure, ΔH° = -890.4 kJ/mol. What will be the amou

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3 years ago

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3 years ago

Read 2 more answers

The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of

gavmur [86]

Answer: The empirical formula for the given compound is $C_3H_6O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Conversion factor: 1 g = 1000 mg

Mass of $CO_2=6.32mg=0.00632g$

Mass of $H_2O=2.58g=0.00258g$

Mass of compound = 2.78 mg = 0.00278 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.00632 g of carbon dioxide, $\frac{12}{44}\times 0.00632=0.00172g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.00258 g of water, $\frac{2}{18}\times 0.00258=0.000286g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.00278) - (0.00172 + 0.000286) = 0.000774 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.00172g}{12g/mole}=1.43\times 10^{-4}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.000286g}{1g/mole}=2.86\times 10^{-4}moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.000774g}{16g/mole}=4.83\times 10^{-5}moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $4.83\times 10^{-5}mol$

For Carbon = $\frac{1.43\times 10^{-4}}{4.83\times 10^{-5}}=2.96\approx 3$

For Hydrogen = $\frac{2.86\times 10^{-4}}{4.83\times 10^{-5}}=5.92\approx 6$

For Oxygen = $\frac{4.83\times 10^{-5}}{4.83\times 10^{-5}}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 6 : 1

Hence, the empirical formula for the given compound is $C_3H_{6}O_1=C_3H_6O$

3 0

3 years ago