Question

A mass of sodium chloride formed when 5.3g of sodium carbonate is dissolved in 250ml of 0.5 molar hcl solution wii be

Answer 1

HELLO THERE!

Let's solve this question step by step:

First, from the molarity of HCl solution, find out the number of moles of HCl.

Then, find out the moles of Sodium carbonate that is equivalent to 5.3 grams of it.

Then, according to the reaction, find out the limiting reagent.

Then, find out the number of moles of NaCl formed. After that, convert it into mass of NaCl (or directly find out the mass of NaCl).

Molecular weight of NaCl = 23 + 35.5 = 58.5

Let's GO!

We know,

Molarity = Number of moles x 1000 / Volume of solution (mL)

Hence, Number of moles = Molarity x Volume / 1000

So, number of moles of HCl:

$\frac{0.5\times250}{1000} = \frac{1}{8}$

Again,

Number of moles = Given mass / Molecular mass

And, molecular mass of sodium carbonate = (2 x 23) + (48 + 12)= 106

So, number of moles of Sodium carbonate taken:

$\frac{0.53}{106} = \frac{1}{20}$

Now, the reaction is:

Na₂CO₃ + 2HCl ------> 2NaCl + H₂O + CO₂

From the reaction, we see that 1 mole of sodium carbonate requires 2 moles of HCl for complete reaction.

So, 1/20 moles of sodium carbonate require =

$2\times \frac{1}{20} = \frac{1}{10}$

moles of HCl.

But, we have 1/8 moles of HCl, which is more than the required amount.

(as 1/10 = 0.1, and 1/8 = 0.125).

So, sodium carbonate will be completely used up, and it is the limiting reagent.

So now,

106 grams of sodium carbonate reacts to give (2 x 58.5) grams of NaCl.

So, 5.3 grams of sodium carbonate will give:

$\frac{117}{106}\times5.3$

= 5.85 grams.

THIS IS YOUR ANSWER...

Thanks!