Which of the following diagrams represents a single replacement reaction?

Please somebody give me the answers

RSB [31]

Answer:

1. 9.4 grams of methane produce 25.85 grams of CO2

2.Grams of water produced = 11.81 grams

3.Mass of Methane produced by 10.1 gram of O2 = 2.52 grams

4.Amount of methane consumed = 46.9 grams

5. Grams of Co2 produced = 8.32 grams

Explanation:

Molar masses :

Methane = CH4 = mass of C + 4x (mass of H)

CH4 = 12 +4(1) = 16 grams

1 mole of CH4 = 16 gram

Oxygen O2 = 2 x (mass of O) = 2x(16) = 32 gram (1 mole of O2 =32 gram)

Carbon Dioxide =CO2 = mass of C + 2(mass of O)

= 12 + 2(16)

= 44 grams (1 mole of CO2 = 44 gram )

Water = H2O = 18 grams ( 1 mole of H2O = 18 gram)

1 mole of each molecule is equal to their molar masses

The balanced equation is :

$1CH_{4}(g)+2O_{2}\rightarrow 1CO_{2}+2H_{2}O(l)$

According to Stoichiometry :

1 mole of CH4 = 2 Mole of O2 = 1 mole of CO2 = 2 mole of H2O

1. From the equation ,

1 mole of methane produce =1 mole of CO2

16 gram of methane = 44 gram of CO2

1 gram of methane =

$\frac{44}{16}$ gram of CO2

9.4 gram of CH4 =

$\frac{44}{16}\times 9.4$ gram of CO2

= 25 .85 gram of CO2

2.

2 mole of O2 produces = 2 mole of H2O(water)

1 mole of O2 produces = 1 mole of H2O

32 gram of O2 = 18 gram of water

1 gram of O2 =

$\frac{18}{32}$

21 gram of O2 =

$\frac{18}{32}\times 21$

11.81 gram of water

3. 1 mole of CH4 = 2 mole of O2

16 gram of CH4 = 2(32) = 64 grams of O2

64 gram of O2 needs = 16 grams of CH4

1 gram of O2 needs =

$\frac{16}{64}$

10.1 gram of O =

$\frac{16}{64}\times 10.1$ of CH4

= 2.52 gram

4.

1 mole of CO2 is produced from = 1 mole of CH4

44 gram of CO2 is produced from 16 gram of CH4

1 gram CO2 =

$\frac{16}{44}$ gram of CH4

129 gram of CO2 =

$\frac{16}{44}\times 129$ gram of CH4

= 46.90 grams

5.

2 mole of O2 produce = 1 mole of CO2

2x 32 gram of O2 = 44 gram of CO2

1 gram of O2 =

$\frac{44}{64}$ of CO2

12.1 gram of O2 produce=

$\frac{44}{64}\times 12.1$ of CO2

= 8.318 gram

Note : Write the quantity give on left side of "="

write the substance asked on right side of "="

8 0

2 years ago

WHAT IS THE STOCK NAME FOR FeSO^3

podryga [215]

Feso3 compound name

Iron(II) Sulfite FeSO3 Molecular Weight

Hope this helps!

Have a great day :)

4 0

2 years ago

I have no idea how to do this and need help

den301095 [7]

Answer:

60.55%

Explanation:

nO3=155/48

nO2 used: 3/2nO3=4.84375

percent yield: 4.84375/8=60.55%

5 0

2 years ago

What is the mass of 0.28 mole of iron?

vredina [299]

Answer:

Your answer should be 15.68 grams.

Explanation:

Seeing as 1 mole has a mass of 56 g, 56*0.28 would get you 15.68 g.

6 0

3 years ago

Calculate the solubility of carbon dioxide at 400 kPa.

BaLLatris [955]

The solubility of carbon dioxide at 400 kPa at room temperature is ;

( B ) 0.61 CO2/L

Given data

pressure of CO₂ = 400 Kpa = 3.95 atm

Kh of CO₂ = 3.3 * 10⁻² mol/L.atm

<h3>Calculate the solubility of carbon dioxide </h3>

Solubility = pressure * Kh value of CO₂

= 3.95 atm * 3.3 * 10⁻² mol / L.atm

= 0.13 mol/l CO₂

= 0.61 CO₂ / L

Hence we can conclude that the solubility of CO₂ at 400 kPa is 0.13 mol/l CO₂.

Learn more about solubility : brainly.com/question/23946616

From the options the closest answer is ( B ) 0.61 CO₂ / L

7 0

2 years ago