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3 years ago

Sodium chloride is produced from its elements

Chemistry

1 answer:

Kisachek [45]3 years ago

6 0

Answer:

Mass of Sodium = 574.75 g

Mass of Chlorine = 886.25 g

Explanation:

The balance chemical equation for the synthesis of NaCl is,

2 Na + Cl₂ → 2 NaCl

Step 1: Find out moles of each reactant required,

According to balance chemical equation,

2 moles of NaCl is produced by = 2 moles of Na

So,

25 moles of NaCl will be produced by = X moles of Na

Solving for X,

X = 25 mol × 2 mol / 2 mol

X = 25 moles of Na

Similarly for Cl₂,

According to balance chemical equation,

2 moles of NaCl is produced by = 1 mole of Cl₂

So,

25 moles of NaCl will be produced by = X moles of Cl₂

Solving for X,

X = 25 mol × 1 mol / 2 mol

X = 12.5 moles of Cl₂

Step 2: Convert each moles to mass as;

Mass = Moles × Atomic Mass

For Na,

Mass = 25 mol × 22.99 g/mol

Mass = 574.75 g

For Cl₂,

Mass = 12.5 mol × 70.90 g/mol

Mass = 886.25 g

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Answer:

pH =1 2.84

Explanation:

First we have to start with the reaction between HCl and KOH:

$HCl~+~KOH->~H_2O~+~KCl$

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$M=\frac{mol}{L}$

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$0.282M=\frac{mol}{0.01L}$

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We can do the same for the KOH values ( $50mL=0.05L$ and $0.141M$ ).

$0.141M=\frac{mol}{0.05L}$

$mol=0.141*0.05$

$mol=0.00705$

So, we have so far 0.00282 mol of HCl and 0.00705 mol of KOH. If we check the reaction we have a molar ratio 1:1, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an excess of KOH. This excess can be calculated if we substract the amount of moles:

$0.00705-0.00282=0.00423mol~of~KOH$

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$60mL=0.06L$

Now we can calculate the concentration:

$M=\frac{0.00423mol}{0.06L}$

$M=0.0705$

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$pOH=-Log(0.0705)$

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