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2 years ago

What volume of a 1.5 M KOH solution is needed to provide 3.0 moles of KOH?

Chemistry

1 answer:

Brums [2.3K]2 years ago

8 0

Answer:

Volume = 2.0 liter of 1.5 M solution of KOH

Explanation:

1) Data:

a) Solution: KOH

b) M = 1.5 M

c) n = 3.0 mol

d) V = ?

2) Formula:

Molarity is a unit of concentration, defined as number of moles of solute per liter of solution:

M = n / V in liter

3) Calculations:

Solve for n: M = n / V ⇒ V = n / M

Substitute values: V = 3.0 mol / 1.5 M = 2.0 liter

You must use 2 significant figures in your answer: 2.0 liter.

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satela [25.4K]

Answer:

$5.6gNa_2SO_4$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2NaOH(aq)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)$

Therefore, since the masses of both of the reactants are given, one computes the available moles of sulfuric acid and those moles of it consumed by the sodium hydroxide as shown below:

$n_{H_2SO_4}^{available}=6.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.0704molH_2SO_4\\n_{H_2SO_4}^{consumed\ by\ NaOH}=3.14gNaOH*\frac{1molNaOH}{40gNaOH}*\frac{1molH_2SO_4}{2molNaOH}=0.04molH_2SO_4$

In such a way, since there is more available sulfuric acid than it that is consumed, the sodium hydroxide is the limiting reagent, consequently, the maximum mass of sodium sulfate turns out:

$m_{Na_2SO_4}=0.04molH_2SO_4*\frac{1molNa_2SO_4}{1molH_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4}=5.6gNa_2SO_4$

Best regards.

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