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Sav [38]
3 years ago
14

How many moles of NaCl will be produced from 83.0g of Na, assuming Cl2 is available in excess?

Chemistry
2 answers:
SashulF [63]3 years ago
5 0

Answer:

2.57

Explanation:

Sindrei [870]3 years ago
3 0

Answer: The number of moles of NaCl produced will be, 3.61 moles.

Explanation : Given,

Mass of Na = 83.0 g

Molar mass of Na = 23 g/mol

First we have to calculate the moles of Na

\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}

\text{Moles of }Na=\frac{83.0g}{23g/mol}=3.61mol

Now we have to calculate the moles of NaCl

The balanced chemical equation is:

2Na+Cl_2\rightarrow 2NaCl

From the reaction, we conclude that

As, 2 mole of Na react to give 2 mole of NaCl

So, 3.61 mole of Na react to give 3.61 mole of NaCl

Therefore, the number of moles of NaCl produced will be, 3.61 moles.

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formic acid buffer containing 0.50 M HCOOH and 0.50 M HCOONa has a pH of 3.77. What will the pH be after 0.010 mol of NaOH has b
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Answer:

pH = 3.95

Explanation:

It is possible to calculate the pH of a buffer using H-H equation.

pH = pka + log₁₀ [HCOONa] / [HCOOH]

If concentration of [HCOONa] = [HCOOH] = 0.50M and pH = 3.77:

3.77 = pka + log₁₀ [0.50] / [0.50]

<em>3.77 = pka</em>

<em />

Knowing pKa, the NaOH reacts with HCOOH, thus:

HCOOH + NaOH → HCOONa + H₂O

That means the NaOH you add reacts with HCOOH producing more HCOONa.

Initial moles of 100.0mL = 0.1000L:

[HCOOH] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOOH

[HCOONa] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOONa

After the reaction, moles of each species is:

0.0500moles HCOOH - 0.010 moles NaOH (Moles added of NaOH) = 0.0400 moles HCOOH

0.0500moles HCOONa + 0.010 moles NaOH (Moles added of NaOH) = 0.0600 moles HCOONa

With these moles of the buffer, you can calculate pH:

pH = 3.77 + log₁₀ [0.0600] / [0.0400]

<h3>pH = 3.95</h3>

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