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padilas [110]
3 years ago
15

Given the following chemical reaction: 2 O2 + CH4 CO2 + 2 H2O What mass of CH4 is required to completely react with 100 grams of

O2? 6 C 12.01 Carbon 1 H 1.01 Hydrogen 8 O 16.00 Oxygen A. 10 grams CH4 B. 20 grams CH4 C. 25 grams CH4 D. 45 grams CH4
Chemistry
1 answer:
AfilCa [17]3 years ago
6 0

Answer:

The mass methane (CH4) required is, 25 grams  (option C)

Explanation:

Mass of oxygen gas = 100 g

Molar mass of oxygen gas = 32 g/mole

Molar mass of methane gas = 16 g/mole

<u>Step 1:</u> Balance the reaction

2O2 + CH4 → CO2 + 2H2O

<u>Step 2:</u> Calculate the moles of O2:

Moles O2 = mass of O2 / Molar mass of o2  

Moles O2 = 100g / (32g/mole) = 3,125 moles

⇒From the balanced reaction we conclude that  2 moles of O2 react with 1 mole of CH4

⇒ So, 3.125 moles of O2 react with 3.125/2 = 1.563 moles of CH4

<u>Step 3:</u> Calculate the mass of CH4:

Mass of CH4 = moles of CH4 x Molar mass of CH4

Mass of CH4 = 1.563 moles / (16g/ moles) = 25.008 grams CH4

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Answer:

80.92, Krypton

Explanation:

<u>What is effusion?</u>

• It is a process where gas escapes through a pinhole (a very small hole) into a region of low pressure or vacuum

<u>Graham's law of effusion of </u><u>gas</u>

• states that at a given constant temperature and pressure, the rate of effusion of gases is inversely proportional to the square root of their molar masses

\boxed{ \frac{Rate_1}{Rate_2} =  \sqrt{ \frac{M_2}{M_1} } }

<u>Calculations</u>

Nitrogen exist as N₂ at room temperature, thus its molar mass is 2(14)= 28.

Let the rate and molar mass of unknown gas be Rate₂ and M₂ respectively.

Since N₂ effuses 1.7 times as fast as the unknown gas,

Rate₁= 1.7(Rate₂)

\frac{Rate_1}{Rate_2} = 1.7

1. 7 =  \sqrt{ \frac{M_2}{28} }

Square both sides:

2.89  = \frac{M_2}{28}

Multiply both sides by 28:

2.89(28)= M₂

M₂= 80.92

<u>Identity of </u><u>gas</u>

The molar mass of 80.92 lies between Bromine and Krypton. However since Bromine exist as Br₂, the value of it's molar mass would be 159.8 instead. Hence, Bromine is eliminated.

If the gas is a diatomic element, the atomic weight is 80.92 ÷2= 40.46. Thus, we are now considering if Argon could be its identity. However, Argon is a noble gas and will not exist as a diatomic element. Argon is therefore eliminated too.

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When 2.94 g of a metal combined with fluorine gas, 7.47 g of a metal fluoride formed. Given this information, calculate the mass
Nutka1998 [239]

 The mass % F  in  the metal fluoride that is formed is  60.64%


 <u><em>calculation</em></u>

Step 1:  calculate the mass  of fluorine (F)

mass  of F =  mass  of  metal fluoride  -  mass of the metal

     = 7.47 g - 2.94 g= 4.53 g

Step 2:  calculate  the  % mass of F

Mass % of  F =  mass  of F / mass  of metal  fluoride  x 100

mass % of F  is therefore  =  4.53  g/ 7.47  x 100 = 60.64%

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