Question

What weight of BaCl2 would react with 25.4 g of Na2SO4 to produce 46.6g BaSO4 and 23.4g of NaCl.​

Answer 1

$\huge \boxed{ \bf{Answer}}$



A/Q,

Mass of Na2O4 = 25.4 g

Mass of BaSO4 = 46.6 g

Mass of NaCl = 23.4 g

Mass of BaCl2 = ?


So, when BaCl2 reacts with Na2SO4.

$\bf{BaCl_{2} + Na_{2}SO_{4} - - - > BaSO_{4} + 2NaCl}$


As, we know that according to law of conservation of mass,

$\underline{ \bf{Total \: mass \: of \: the \: reactants = Total \: mass \: of \: the \: products.}}$


Now,

Let, the mass of BaCl2 be 'x'.

$\therefore \: (x + 25.4) \: = \: (46.6 \: + \: 23.4)$

$= > \: x \: + \: 25.4 \: = 70$

$= > \: x \: = \: 70 \: - \: 25.4$

$= > \: x = 44.6 \: g$


So, our required mass of

$\boxed{ \underline{\bf{BaCl_{2} \: = \: 44.6 \: g}}}$

Answer 2

Bacl2+Na2So4=>BaSo4+NaCl

BaCl2 + Na2So4 => BaSo4 + 2 NaCl

44.6g+25.4g=>46.6g+23.4g