100 mL of 20.8% $BaCl_{2}$ solution and 50 mL of 9.8% $H_{2}SO_{4}$ Solution will form $BaSO_{4}$ $(Ba = 137,Cl = 35.5,S = 32,H = 1,0 = 16)$ $BaCl_{2} + H_{2}SO_{4} \longrightarrow Ba_{2}SO_{4} + 2HCl$

Question

Dr. Abubacker · Accepted Answer

(b)
$BaCl_{2} + H_{2}SO_{4} \longrightarrow BaSO_{4} + H_{2}O$
208 g 98g 233 g
100 mL of 20.8% $BaCl_{2}$ solution contains $= 20.8\ g\ BaCl_{2}$
$50\$mL of 9.8% $H_{2}SO_{4}$ solution contains $= 4.9\ g\ H_{2}SO_{4}$
Here, $H_{2}SO_{4}$ is the limiting reactant.
$\because 98g\ H_{2}SO_{4}$ gives $BaSO_{4} = 233\ g$
$	herefore$ 4.9 g $H_{2}SO_{4}$ will give $BaSO_{4} = \frac{233 	imes 4.9}{98}$

Q. 100 mL of 20.8% $BaCl_{2}$ solution and 50 mL of 9.8% $H_{2}SO_{4}$ Solution will form $BaSO_{4}$ $(Ba = 137,Cl = 35.5,S = 32,H = 1,0 = 16)$ $BaCl_{2} + H_{2}SO_{4} \longrightarrow Ba_{2}SO_{4} + 2HCl$

Explanation

Reviewed by Dr. Abubacker

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