Question

bacl2 is treated with na3po4 to form precipitate of barium phosphate . the equivalent mass of barium phosphate in the reaction will be​

Answer 1

Answer:

601 g

Explanation:

3 Bacl2 + 2Na3PO4 -> Ba3(PO4)2 + 6Nacl

Answer 2

Concept:

Equivalent mass is the number of substances in a reaction that totally react with one another.

Given:

$BaCl_{2}$  treated with $Na_{3} PO_{4}$.

Find:  Calculate the equivalent mass of barium phosphate in the reaction?

Solution:

The equivalent mass can be calculated as:

                         Equivalent mass = $\frac{Molar mass}{Valency}$

The given reaction is as follows below:

                   $3BaCl_{2}$  + $2Na_{3} PO_{4}$ → $Ba_{3} (PO_{4})_{2}$ + $6NaCl$

The valency of barium in barium phosphate is 2.

The molar mass of barium phosphate = 601.93 amu

By substituting the given values in the equivalent mass expression, we get,

                        Equivalent mass = $\frac{601.93 amu}{2}$ = 300.965 amu

Hence, when $BaCl_{2}$ treated with $Na_{3} PO_{4}$ to form a precipitate of barium phosphate, the equivalent mass of barium phosphate in the reaction will be​ 300.965 amu.

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