Question

Calculate the moles of NaOH required to neutralize the solution produced by dissolving 1.1 g P4O6 in water. Use the following reactions: $P_{4}O_{6}+6H_{2}O \longrightarrow 4H_{3}PO_{3}$ $2NaOH+H_{3}PO_{4}\longrightarrow Na_{2}HPO_{3}+2H_{2}O$

Answer 1

"$P_{ 4 }O_{ 6 }\quad +\quad 6H_{ 2 }O \rightarrow \quad 4H_{ 3 }PO_{ 3 }$

Molar mass of $P_{ 4 }O_{ 6 } = 220$

Product formed by $1 mol P_{ 4 }O_{ 6 }$is neutralised by 8 mol NaOH.

Given, dissolved 1.1 g $P_{ 4 }O_{ 6 }$in water.

Therefore, the Product formed by $\frac {1.1}{220} mol P_{ 4 }O_{ 6 }$will be neutralised by$\frac { 1.1 }{ 220} \ times 8 mol$ NaOH

Hence, Molarity of NaOH solution is 0.1M

0.1 mol NaOH is present in 1 L solution.

Therefore $\frac { 1.1 }{ 220 } \quad \times \quad 8\quad mol\quad NaOH\quad is\quad present\quad in\quad \frac { 1.1\quad \times \quad 8 }{ 220\quad \times \quad 0.1 } L\quad =\quad \frac { 88 }{ 220 } L\quad =\quad \frac { 4 }{ 10 } L\quad =\quad 0.4L$

= 400 ml of NaOH solution

Thus, the volume of  NaOH = 0.4 L = 400 ml"

Answer 2

Answer:

4 ÷ 10² moles

Explanation:

From these two reactions we can understand that the no of moles ratio between P406 AND NaOH is 1:8

No of moles of P4O6 is 1.1/220 = 5 / 10³ moles.

No of moles of NaOH is now 8 × 5 / 10³  = 4 ÷ 10² moles

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