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lidiya [134]
3 years ago
9

How many milliliters of ammonium sulfate solution having a concentration of 0.218 M are needed to react completely with 62.6 ml

of 2.31 M sodium hydroxide solution?
Chemistry
1 answer:
BartSMP [9]3 years ago
6 0

Answer:

330 mL of (NH₄)₂SO₄ are needed

Explanation:

First of all, we determine the reaction:

(NH₄)₂SO₄  +  2NaOH →  2NH₃  +  2H₂O  +  Na₂SO₄

We determine the moles of base:

(First, we convert the volume from mL to L) → 62.6 mL . 1L/1000 mL = 0.0626L

Molarity . volume (L) = 2.31 mol/L . 0.0626 L = 0.144 moles

Ratio is 2:1. Therefore we make a rule of three:

2 moles of hydroxide react with 1 mol of sulfate

Then, 0.144 moles of NaOH must react with (0.144 .1) /2 = 0.072 moles

If we want to determine the volume → Moles / Molarity

0.072 mol / 0.218 mol/L = 0.330 L

We convert from L to mL → 0.330L . 1000 mL/1L = 330 mL

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3 years ago
A 36.165 mg36.165 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion an
erma4kov [3.2K]

Answer:

The empirical formula for the compound is C6H12SO2.

Explanation:

We'll begin by writing out what was given from the question. This is shown:

Let us consider the First experiment:

Mass of the compound = 36.165 mg

Mass of CO2 = 64.425 mg

Mass of H2O = 26.373 mg

Data obtained from the Second experiment:

Mass of compound = 47.029 mg

Mass of SO2 = 20.32 mg

Next, we'll determine the mass of C, H and S. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 64.425 Mass of C = 17.57 mg

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 26.373

Mass of H = 2.93 mg

Molar Mass of SO2 = 32 + (16x2) = 64g/mol

Mass of S in SO2 = 32/64 x 20.32

Mass of S = 10.16 mg

At this stage, it is important we determine the percentage composition of C, H, S and O. This is illustrated below:

% of C = 17.57/36.165 x 100 = 48.58%

% of H = 2.93/36.165 x 100 = 8.10%

% of S = 10.16/47.029 x 100 = 21.60%

% of O = 100 - (48.58 + 8.1 + 21.6)

% of O = 21.72%

Now we can easily obtain the empirical formula for the compound by doing the following.

Step 1:

Divide by their molar mass

C = 48.58/12 = 4.0483

H = 8.10/1 = 8.1

S = 21.60/32 = 0.675

O = 21.72/16 = 1.3575

Step 2:

Divide by the smallest:

C = 4.0483/0.675 = 6

H = 8.1/0.675 = 12

S = 0.675/0.675 = 1

O = 1.3575/0.675 = 2

From the calculations made above, empirical formula for the compound is C6H12SO2

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5. A sample of benzene (C6H6), weighing 7.05 g underwent combustion in a bomb calorimeter by the following reaction:
Cloud [144]

Answer:

A sample of benzene (C6H6), weighing 7.05 g underwent combustion in a bomb calorimeter by the following reaction:

2 C6H6 (l) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (l)

The heat given off was absorbed by 500 g of water and caused the temperature of the water and the calorimeter to rise from 25.00 to 53.13 oC. The heat capacity of water = 4.18 J/g/oC and the heat capacity of the calorimeter = 10.5 kJ/oC. (1) what is the ΔH of the reaction?

Explanation:

The heat energy released by the reaction = heat absorbed by calorimeter + heat absorbed by water

Heat absorbed by water = mass of water x specific heat capacity of water x change in temperature

Heat absorbed by water =  500 g x 4.18 J/g. oC x (53.13-25.00)oC

                                         = 58791.7 J

Heat absorbed by calorimeter = heat capacity of calorimeter x change in temperature

Heat absorbed by calorimeter = 10.5 x 10^3 J /oC  x (53.13-25.00)oC

                                                  =295365 J

Total heat energy absorbed = 58791.7 J + 295365 J  = 354156.7 J

Number of moles of benzene given is:

number of moles = goven mass of benzene /its molar mass

=7.05 g / 78.0 g/mol

=0.0903mol

Hence, the heat released by the reaction is:

= 354156.7 J / 0.0903 mol

=  3922.00 kJ/mol

Answer:

The heat released during the combustion of 7.05g of benzene is 3922.00kJ/mol.                                              

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3 years ago
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