Question

2 K3PO4 + 3 BaCl2 Ba3(PO4)2 + 6 KCl

Calculate the mass of BaCl2 that is required to react with 21.2 grams K3PO4 of in the above reaction

Answer 1

Answer:

The mass of $BaCl_2$ that is required to react with $21.2$ g of  $K_3PO_4$  in the given reaction is  $31.19$ g·

Explanation:

The balanced chemical equation that is given is,

    $2 \ K_3PO_4\ +\ 3\ BaCl_2\ --->\ Ba_3(PO_4)_2\ +\ 6\ KCl$

From the reaction, it is clear that,

 Two molecules of $K_3PO_4$ reacted with three molecules of $BaCl_2$ to get one molecule of $Ba_3(PO_4)_2$ and six molecules of $KCl$ as products·

Given that,

The weight of $K_3PO_4$ taken $=\ 21.2$ g

We know that,

The molecular mass of  $K_3PO_4\ =\ 212.27$ g

The molecular mass of $BaCl_2\ =\ 208.23$ g

Two molecules of $K_3PO_4$  $=\ 2\ *\ 212.27\ g\ =\ 424.54$  g

Three molecules of $BaCl_2$  $3\ *\ 208.23\ g\ =\ 624.69$ g

So,

    $424.54$ g of $K_3PO_4$ needs $624.69$ g of $BaCl_2$

Then,

     $21.2$ g of  $K_3PO_4$  requires $\frac{21.2\ g \ *\ 624.69\ g}{424.54\ g}$

⇒   $31.19$ g of  $BaCl_2$

Hence,

The mass of $BaCl_2$ that is required to react with $21.2$ g of  $K_3PO_4$ in the above - balanced chemical reaction is  $31.19$ g ·