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lana [24]
3 years ago
13

Balance the following equation. Then, given the moles of reactant or product below, determine the corresponding amount in moles

of each of the other reactants and products.NH3+O2⟶N2+H2O
a.4molNH3b.4 molN2c. 4.5molO2
Chemistry
1 answer:
Mrrafil [7]3 years ago
6 0

Answer:

Option a → 4 mol NH₃

Explanation:

This the unbalanced reaction

NH₃  +  O₂  ⟶  N₂  +  H₂O

The balanced reaction:

4NH₃  +  3O₂  →  2N₂  +  6H₂O

4 mol of ammonia

3 mol of oxygen

2 mol of nitrogen

6 mol of water

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A gas mixture contains twice as many moles of o2 as n2. addition of 0.200 mol of argon to this mixture increases the pressure fr
pantera1 [17]
Number of moles of oxygen = x

number of moles of nitrogen = y

x = 2y

initial pressure, p1 = 0.8 atm

final pressure, p2 = 1.10 atm

At constant volume and temperature p1 / n1 = p2 / n2

=> p1 / p2 = n1 / n2

n1 = x + y = 2y + y = 3y

n2 = 0.2 + 3y

=> p1 / p2 = 3y / (0.2 + 3y)

=> 0.8 / 1.10 = 3y / (0.2 + 3y)

=> 0.8 (0.2 + 3y) = 1.10 (3y)

0.16 + 2.4y = 3.3y

=> 3.3y - 2.4y = 0.16

=> 0.9y = 0.16

=> y = 0.16 / 0.9

=. x = 2*0.16/0.9 = 0.356

Answer: 0.356 moles O2
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3 years ago
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Temperature of somewhere.
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Which scenario is an example of primary succession?
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8 0
3 years ago
Please somebody give me the answers
RSB [31]

Answer:

1. 9.4 grams of methane produce<u> 25.85</u> grams of CO2

2.Grams of water produced = <u>11.81 grams</u>

3.Mass of Methane produced by 10.1 gram of O2 = <u>2.52 grams</u>

4.Amount of methane consumed = <u>46.9 grams</u>

5. Grams of Co2 produced =<u> 8.32 grams</u>

<u></u>

Explanation:

Molar masses :

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

CH4 = 12 +4(1) = 16 grams

<u>1 mole of CH4 = 16 gram</u>

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 <u>(1 mole of CO2 = 44 gram )</u>

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
3 years ago
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