Question

To prepare common salt, 26.5g Na2CO3 is mixed with 36.5g of HCI. Which reactant and how much will be left over after reaction?

a) 26.5 g Na2CO3
b) 18.25g HCI
c) 8.25g HCI
d) 53g Na2CO3​

Answer 1

Answer :-

18.25 grams of HCl will be left over after reaction. [Option.b]

Explanation :-

The balanced chemical equation for the reaction is :-

Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O

Moles of Na₂CO₃ :-

= Given Mass/Molar mass

= 26.5/106

= 0.25 mole

Moles of HCl :-

= Given Mass/Molar mass

= 36.5/36.5

= 1 mole

______________________________

From the balanced equation, we can understand that 1 mole of Na₂CO₃ combines with 2 moles of HCl.

In this case, 0.25 mole of Na₂CO₃ will combine with 0.5 mole of HCl. So, (1 - 0.5) = 0.5 mole of HCl will be left over after reaction .

Mass of HCl left over :-

= Number of moles × Molar mass

= 0.5 × 36.5

= 18.25 g

Answer 2

Answer:

Given :-

• To prepare common salt, 26.5 g Na₂CO₃ is mixed with 36.5 g of HCl.

To Find :-

• How much will be left over after the reaction.

Solution :-

$\clubsuit$ Balanced Equation :

$\bigstar\: \: \sf\boxed{\bold{Na_2CO_3 + 2HCL \longrightarrow\: 2NaCl + CO_2 + H_2O}}$

First, we have to find the moles of Na₂CO₃ and HCl :

As we know that,

$\clubsuit$ Moles Formula :

$\mapsto \sf\boxed{\bold{\pink{Moles =\: \dfrac{Mass}{Molar\: Mass}}}}$

${\small{\bold{\purple{\underline{\leadsto\: In\: case\: of\: Na_2CO_3\: :-}}}}}$

Given :

• Mass = 26.5 g
• Molar Mass = 106

According to the question by using the formula we get,

$\implies \sf Moles_{(Na_2CO_3)} =\: \dfrac{26.5}{106}$

$\implies \sf Moles_{(Na_2CO_3)} =\: \dfrac{\dfrac{265}{10}}{106}$

$\implies \sf Moles_{(Na_2CO_3)} =\: \dfrac{265}{10} \times \dfrac{1}{106}$

$\implies \sf Moles_{(Na_2CO_3)} =\: \dfrac{265}{1060}$

$\implies \sf\bold{\green{Moles_{(Na_2CO_3)} =\: 0.25\: moles}}$

${\small{\bold{\purple{\underline{\leadsto\: In\: case\: of\: HCl\: :-}}}}}$

Given :

• Mass = 36.5 g
• Molar Mass = 36.5

According to the question by using the formula we get,

$\implies \sf Moles_{(HCl)} =\: \dfrac{36.5}{36.5}$

$\implies \sf Moles_{(HCl)} =\: \dfrac{\dfrac{365}{10}}{\dfrac{365}{10}}$

$\implies \sf Moles_{(HCl)} =\: \dfrac{365}{10} \times \dfrac{10}{365}$

$\implies \sf Moles_{(HCl)} =\: \dfrac{\cancel{3650}}{\cancel{3650}}$

$\implies \sf\bold{\green{Moles_{(HCl)} =\: 1\: moles}}$

Now,

We have :

• Moles of Na₂CO₃ = 0.25 moles
• Moles of HCl = 1 moles

In this case, we have to combine 1 moles of Na₂CO₃ with 2 moles of HCl we get,

$\implies \sf 1 - 0.5\: moles$

$\implies \sf\bold{0.5\: moles}$

Hence, the required mass of HCl left over after reaction is :

$\longrightarrow \sf\bold{\pink{Mass\: of\: HCl =\: Total\: number\: of\: moles \times Molar\: mass}}$

$\longrightarrow \sf Mass\: of\: HCl =\: 0.5 \times 36.5$

$\longrightarrow \sf Mass\: of\: HCl =\: \dfrac{5}{10} \times \dfrac{365}{10}$

$\longrightarrow \sf Mass\: of\: HCl =\: \dfrac{1825}{100}$

$\longrightarrow \sf\bold{\red{Mass\: of\: HCl =\: 18.25\: g\: of\: HCl}}$

$\therefore$ The mass of HCl that left over after reaction is 18.25 g of HCl.

Hence, the correct options is option no (b) 18.25 g of HCl.