Calculate number of moles of ${N a}_{2} {S O}_{4}$ produced from 1 mole of $N a O H$
$2 N a O H + H_{2} {S O}_{4} \to {N a}_{2} {S O}_{4} + 2 H_{2} O$

Open in App

Solution

$2 N a O H + H_{2} S O_{4} ⟶ {N a}_{2} S O_{4} + 2 H_{2} O$
From the above reaction,
1 mole of ${N a}_{2} S O_{4}$ is produced by 2 moles of $N a O H$
$∴$ No. of moles of ${N a}_{2} S O_{4}$ produced by 1 mole of $N a O H = \frac{1}{2} moles$
Hence, half mole of ${N a}_{2} S O_{4}$ nis produced by 1 mole of $N a O H$ .

Suggest Corrections

Similar questions

N a_{2} S O_{4}

2 N a O H + H_{2} S O_{4} \to N a_{2} S O_{4} + 2 H_{2} O

H_{2} {S O}_{4}

(a) H_{2} {S O}_{4} + N a O H \to N a H {S O}_{4} + H_{2} O

(b) H_{2} {S O}_{4} + 2 N a O H \to {N a}_{2} {S O}_{4} + 2 H_{2} O

N a O H + H C l (d i l u t e d) \to N a C l + H_{2} O

2 N a C l + H_{2} S O_{4} \to N a_{2} S O_{4} + 2 H C l

Na + O_{2} \underset{}{\to} {Na}_{2} O

H_{2} O_{2} \underset{}{\to} H_{2} O + O_{2}

Mg (OH)_{2} + HCI \overset{}{\to} {MgCl}_{2} + H_{2} O

Fe + O_{2} \underset{}{\to} {Fe}_{2} O_{3}

Al (OH)_{3} \underset{}{\to} {Al}_{2} O_{3} + H_{2} O

{NH}_{3} + CuO \underset{}{\to} Cu + N_{2} + H_{2} O

{Al}_{2} ({SO}_{4})_{3} + NaOH \underset{}{\to} Al (OH)_{3} + {Na}_{2} {SO}_{4}

{HNO}_{3} + Ca (OH)_{2} \underset{}{\to} Ca ({NO}_{3})_{2} + H_{2} O

NaOH + H_{2} {SO}_{4} \underset{}{\to} {Na}_{2} {SO}_{4} + H_{2} O

{BaCl}_{2} + H_{2} {SO}_{4} \underset{}{\to} {BaSO}_{4} + HCl

Join BYJU'S Learning Program