Question

define limiting reagent. calculate the mass of sodium acetate (CH3COONa) required to make 500 ml of 0.375 molar aqeous solution. Molar mass of sodium acetate is 82.02 gmo1-1​

Answer 1

Answer:

A limiting reagent is a reactant that gets consumed in a chemical reaction first·

The mass of sodium acetate required to make $500$ mL of $0.375$ molar aqueous solution is  $16.4$ g·

Explanation:

A limiting reagent is a reactant that gets consumed in a chemical reaction first· So, the formation of the product gets limited due to this and the reaction will complete faster·

For example,

In a reaction,

$N_2\ +\ 3\ H_2\ --->\ 2\ NH_3$

One mole of nitrogen needs three moles of hydrogen to form two moles of ammonia·

But in the presence of three moles of hydrogen, if one mole is given then the hydrogen becomes the limiting reagent which limits the formation of ammonia in the reaction·

For the second question,

Given that,

Molarity of the aqueous solution $=$ $0.375$ Molar

The volume of the solution  $=$ $500$ mL  $=$ $0.5$ L

The molar mass of sodium acetate  $=$  $82.02\ \frac{g}{mol}$

To calculate the mass of sodium acetate required to make $500$ mL of $0.375$ molar aqueous solution first, we need to find out the number of moles of sodium acetate present in the solution·

For that, we know

      Molarity  $=$  $\frac{No\ Of\ Moles}{Volume\ of\ The\ Solution_{(L)} }$

Therefore,

Number of moles of sodium acetate$=$ Molarity × Volume of the solution in L

On substituting the values we get,

 ⇒  Number of moles of sodium acetate   $=$ $0.375$ M × $0.5$ L

 ⇒  No Of Moles of sodium acetate   $=$ $0.2$ mol

So, we get the number of moles of sodium acetate as $0.2$ mol·

Now to calculate the required mass of sodium acetate,

We know that,

  No Of Moles $=$ $\frac{Required\ Mass}{Molar\ Mass}$

Therefore,

 Required Mass $=$  No Of Moles × Molar Mass

On substituting the values we get,

 ⇒  Required Mass of sodium acetate $=$  $0.2$ mol ×  $82.02\ \frac{g}{mol}$

  ⇒ Required Mass of sodium acetate $=$  $16.4$ g

Therefore, the mass of sodium acetate required to make $500$ mL of $0.375$ molar aqueous solution is  $16.4$ g·