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rjkz [21]
4 years ago
5

What volume (mL) of 0.125 M NaOH is required to neutralize 12.4 mL of 0.369 M HCl? What volume (mL) of 0.125 M NaOH is required

to neutralize 12.4 mL of 0.369 M HCl? 36.6 4.20 0.572 0.000572 0.0273
Chemistry
1 answer:
Darya [45]4 years ago
8 0

Answer:

There is needed 36.6 mL of 0.125M NaOH to neutralize the 12.4 mL 0.369 M HCl

Explanation:

<u>Step 1</u>: The balanced equation

NaOH + HCl ⇔ NaCl + H2O

Sodium hydroxide (NaOH) and hydrochloric acid (HCl) react in a  1:1 mole ratio to produce aqueous sodium chloride and water.

This means 1 mole of NaOH is needed when 1 mole of HCl is consumed to produce 1 mole of NaCl and 1 mole of water.

<u>Step 2</u>: Calculate The required volume NaOH to neutralize  HCl

Because the mole ratio is 1:1 we will use the following formula

C1*V1 = C2*V2

with C1 = the concentration of NaOH = 0.125 M

with V1 = the volume of NaOH = TO BE DETERMINED

with C2 = the concentration of HCl = 0.369M

with V2 = the volume of HCl = 12.4 mL = 0.0124 L

0.125M * V1 = 0.369M * 0.0124 L

V1 = (0.369 * 0.0124)/ 0.125M = 0.0366 L = 36.6 mL

There is needed 36.6 mL of 0.125M NaOH to neutralize the 12.4 mL 0.369 M HCl

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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 "="

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