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eimsori [14]
3 years ago
6

Find the volume of 0.100M hydrochloric acid necessary to react completely with 1.51g Al(OH)3.

Chemistry
1 answer:
shtirl [24]3 years ago
3 0
Reaction equation:
Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O
Moles of Al(OH)₃:
moles = mass/Mr
= 1.51 / (27 + 17 x 3)
= 0.019
Molar ratio Al(OH)₃ : HCl = 1 : 3
Moles of HCl required = 0.019 x 3
=0.057
concentration = moles/volume
volume = 0.057 / 0.1
= 0.57 dm³
= 570 ml
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How many milliliters of 0.200 M FeCl3 are needed to react with an excess of Na2S to produce 0.345 g of Fe2S3 if the percent yiel
GuDViN [60]

Answer:

25.53mL of 0.200 M FeCl₃ are needed to produce 0.345g of Fe₂S₃

Explanation:

Based on the reaction of the problem, 1 mole of Fe₂S₃ is produced from 2 moles of FeCl₃.

0.345g of Fe₂S₃ are (Molar mass: 207.9g/mol):

0.345g of Fe₂S₃ ₓ (1 mol / 207.9g) = <em>1.6595x10⁻³ moles Fe₂S₃</em>

Moles of Fe needed to produce these moles of Fe₂S₃ are:

1.6595x10⁻³ moles Fe₂S₃ ₓ ( 2 moles FeCl₃ / 1 mole Fe₂S₃) =

<em>3.3189x10⁻³ moles of FeCl₃</em>

As the percent yield of the reaction is 65.0%, the moles of FeCl₃ you need to add are:

3.3189x10⁻³ moles of FeCl₃ ₓ (100.0% / 65.0%) = <em>5.106x10⁻³ moles of FeCl₃</em>

A solution 0.200M contains 0.200 moles per L. Volume to obtain 5.106x10⁻³ moles is:

<em>5.106x10⁻³ moles of FeCl₃ ₓ ( 1L / 0.200mol) = 0.02553L = </em>

<h3>25.53mL of 0.200 M FeCl₃ are needed to produce 0.345g of Fe₂S₃</h3>

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3 years ago
Consider that the room is a closed system. Ryan uses the chemical energy in his muscles to create the energy of motion. He uses
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A solution is 40.00% by volume benzene (C6H6) in carbon tetrachloride at 20°C. The vapor pressure of pure benzene at this temper
finlep [7]

Answer:

The total vapor pressure is 84.29 mmHg

Explanation:

Step 1:  Data given

Solution = 40.00 (v/v) % benzene in CCl4

Temperature = 20.00 °C

The vapor pressure of pure benzene at 20.00 °C = 74.61 mmHg

Density of benzene is 0.87865 g/cm3

The vapor pressure of pure carbon tetrachloride is 91.32 mmHg

We suppose the total volume = 100 mL

Step 2: Calculate volume benzene and CCl4

40 % benzene = 40 mL

60 % mL CCl4 = 60 mL

Step 3: Calculate mass benzene

Mass = density * volume

Mass of benzene = 40.00 mL *  0.87865 g/mL = 35.146 g

Step 4: Calculate moles of benzene

Moles = mass / molar mass

Number of moles of benzene  = 35.146 grams / 78 g/mol  = 0.45059 mol

Step 5: Calculate mass of CCl4

Mass of CCl4 = 60 mL * 1.5940 g/mL = 95.64 g

Step 6: Calculate moles CCl4

Number of moles of CCl4 = 95.64 grams / 154g/mol = 0.62104 mol

Step 7: Calculate total number of moles

Total number of moles = moles benzene + moles CCl4

0.45059 moles + 0.62104 moles = 1.07163 mol

Step 8: Calculate mole fraction benzene and CCl4

Mole fraction = moles benzene / total moles

Mole fraction of benzene = 0.45059 / 1.07163 = 0.4205

Mole fraction of CCl4 = 0.62104 / 1.07163 = 0.5795

Step 9: Calculate partial pressure

Partial pressure of benzene = 0.4205 * 74.61 = 31.37 mmHg

Partial pressure of CCl4      = 0.5795 * 91.32 = 52.92 mmHg

Total vapor pressure = 31.37 + 52.92 = 84.29 mmHg

The total vapor pressure is 84.29 mmHg

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