Question

For the following reaction, 63.2 grams of barium hydroxide are allowed to react with 41.2 grams of sulfuric acid .What is the maximum amount of barium sulfate that can be formed? What is the FORMULA for the limiting reagent?What amount of the excess reagent remains after the reaction is complete? 

Answer 1

Ba (OH)$_{2(aq)}+$ H$_2$ SO$_{4(aq)}\to$ BaSO$_{4(s)}+2$H$_2O_{(l)}$

Moles of Ba(OH)₂

$=\frac{63.2}{171.34}=0.3689$

Moles of H₂SO₄

$=\frac{41.2}{98.079}=0.4201$

Limiting reagent is Ba(OH)₂

Mass of BaSO₄ $=0.3689×233.38$

$=86.09g$

Excess H₂SO₄ $=0.4201-0.3689$

$=0.0512\>moles$

$=5.0216g$