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3 years ago

2) What is the concentration of NaCl in a solution prepared by diluting 56.98 ml of 0.5894 M stock

Chemistry

1 answer:

RSB [31]3 years ago

3 0

Answer:

The NaCl concentration will be 0.03 M.

Explanation:

Given data:

Initial volume = V₁ = 56.98 mL (56.98/1000 = 0.05698 L)

Initial concentration = M₁= 0.5894 M

Final volume = V₂= 1.20 L

Final concentration = M₂= ?

Solution:

By diluting the solution volume of solution will increase while number of moles of solute remain the same.

Formula:

Initial concentration × Initial volume = Final concentration × Final volume

M₁V₁ = M₂V₂

M₂ = M₁V₁ / V₂

M₂ = 0.5894 M × 0.05698 L / 1.20 L

M₂ = 0.0336 M /1.20

M₂ = 0.03 M

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Which reactant will be used up first if 78.1g of o2 is reacted with 62.4g of c4h10?

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Answer:

Reagent O₂ will be consumed first.

Explanation:

The balanced reaction between O₂ and C₄H₁₀ is:

2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

Then, by reaction stoichiometry, the following amounts of reactants and products participate in the reaction:

C₄H₁₀: 2 moles
O₂: 13 moles
CO₂: 8 moles
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H: 1 g/mole
O: 16 g/mole

The molar mass of the compounds that participate in the reaction is:

C₄H₁₀: 4*12 g/mole + 10*1 g/mole= 58 g/mole
O₂: 2*16 g/mole= 32 g/mole
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H₂O: 2*1 g/mole + 16 g/mole= 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of reactants and products participate in the reaction:

C₄H₁₀: 2 moles* 58 g/mole= 116 g
O₂: 13 moles* 32 g/mole= 416 g
CO₂: 8 moles* 44 g/mole= 352 g
H₂O: 10 moles* 18 g/mole= 180 g

If 78.1 g of O₂ react, it is possible to apply the following rule of three: if by stoichiometry 416 g of O₂ react with 116 g of C₄H₁₀, 62.4 g of C₄H₁₀ with how much mass of O₂ do they react?

$mass of O_{2} =\frac{416grams of O_{2}*62.4 grams ofC_{4}H_{10} }{116 grams of C_{4}H_{10}}$

mass of O₂= 223.78 grams

But 21.78 grams of O₂ are not available, 78.1 grams are available. Since you have less mass than you need to react with 62.4 g of C₄H₁₀, reagent O₂ will be consumed first.

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