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gulaghasi [49]
2 years ago
5

How many grams of potassium benzoate trihydrate, KC H302 - 3H20. are needed to prepare 1.5 L of a 0.120M solution of potassium b

enzoate? Mass = g
Chemistry
1 answer:
Annette [7]2 years ago
8 0

<u>Answer:</u> The mass of potassium benzoate is 8.40\times 10^{-4}g

<u>Explanation:</u>

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

We are given:

Molarity of solution = 0.120 M

Molar mass of potassium benzoate trihydrate = 214.26 g/mol

Volume of solution = 1.5 L

Putting values in above equation, we get:

0.120M=\frac{\text{Mass of potassium benzoate trihydrate}}{214.26\times 1.5}\\\\\text{Mass of potassium benzoate trihydrate}=\frac{0.120mol/L\times 1.5L}{214.26g/mol}=8.40\times 10^{-4}g

Hence, the mass of potassium benzoate is 8.40\times 10^{-4}g

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Answer:

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Answer these please ASAP need help no idea how to do these
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Answer:

Explanation:

Cu:

Number of moles = Mass / molar masa

2 mol = mass / 64 g/mol

Mass = 128 g

Mg:

Number of moles = Mass / molar masa

0.5 mol = mass / 24 g/mol

Mass =  g

Cl₂:

Number of moles = Mass / molar masa

Number of moles  = 35.5 g / 24 g/mol

Number of moles = 852 mol

H₂:

Number of moles = Mass / molar mass

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Mass =  16 g

P₄:

Number of moles = Mass / molar masa

2 mol  =  mass / 124 g/mol

Mass = 248 g

O₃:

Number of moles = Mass / molar masa

Number of moles  = 1.6 g /48  g/mol

Number of moles = 0.033 mol

H₂O

Number of moles = Mass / molar masa

Number of moles  = 54 g / 18 g/mol

Number of moles = 3 mol

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Number of moles = Mass / molar masa

2 mol  =  mass / 124 g/mol

Mass = 248 g

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Number of moles = Mass / molar masa

Number of moles  = 8.5 g / 17 g/mol

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a)

Given data:

Mass of iron(III)oxide needed = ?

Mass of iron produced = 100 g

Solution:

Chemical equation:

F₂O₃ + 3CO    →    2Fe  + 3CO₂

Number of moles of iron:

Number of moles = mass/ molar mass

Number of moles = 100 g/ 56 g/mol

Number of moles = 1.78 mol

Now we compare the moles of iron with iron oxide.

                        Fe          :           F₂O₃                

                           2          :             1

                          1.78       :        1/2×1.78 = 0.89 mol

Mass of  F₂O₃:

Mass = number of moles × molar mass

Mass = 0.89 mol × 159.69 g/mol

Mass = 142.124 g

100 g of iron is 1.78 moles of Fe, so 0.89 moles of F₂O₃ are needed, or 142.124 g of iron(III) oxide.

b)

Given data:

Number of moles of Al = 0.05 mol

Mass of iodine = 26 g

Limiting reactant = ?

Solution:

Chemical equation:

2Al + 3I₂   →  2AlI₃

Number of moles of iodine = 26 g/ 254 g/mol

Number of moles of iodine = 0.1 mol

Now we will compare the moles of Al and I₂ with AlI₃.

                          Al            :         AlI₃    

                          2             :           2

                         0.05         :        0.05

                           I₂            :         AlI₃

                           3            :          2

                         0.1           :           2/3×0.1 = 0.067

Number of moles of AlI₃ produced by Al are less so it will limiting reactant.

Mass of AlI₃:                            

Mass = number of moles × molar mass

Mass = 0.05 mol × 408 g/mol

Mass = 20.4 g

26 g of iodine is 0.1 moles. From the equation, this will react with 2 moles of Al. So the limiting reactant is Al.

c)

Given data:

Mass of lead = 6.21 g

Mass of lead oxide = 6.85 g

Equation of reaction = ?

Solution:

Chemical equation:

2Pb + O₂   → 2PbO

Number of moles of lead = mass / molar mass

Number of moles = 6.21 g/ 207 g/mol

Number of moles = 0.03 mol

Number of moles of lead oxide = mass / molar mass

Number of moles = 6.85 g/ 223 g/mol

Number of moles = 0.031 mol

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                2          :         1

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Mass = number of moles × molar mass

Mass = 0.015 mol × 32 g/mol

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