In a particular experiment- 310 g of phosphorous P4 reacted with excess of oxygen to form P4O10 in 75Consider only two places after decimal for calculation

P_4 + 5O_2 → P_4O_10 P_4O_10+6H_2O → 4H_3PO_4 In the firest step, for 75 = moles of P_4 ×75100 = 310124×75100 = 1.875 moles Again, P_4O_10 + 6H_2O → 4H_3PO_4 ...

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Standard XIII

Chemistry

Stoichiometric Calculations

In a particul...

Question

In a particular experiment, $310 g$ of phosphorous ( $P_{4}$ ) reacted with excess of oxygen to form $P_{4} O_{10}$ in 75% yield. Then $P_{4} O_{10}$ formed reacted with $54$ g of water to give $H_{3} P O_{4} .$ Amount of $H_{3} P O_{4}$ obtained (in gm) is :
(Consider only two places after decimal for calculation)

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Solution

$P_{4} + 5 O_{2} \to P_{4} O_{10}$
$P_{4} O_{10} + 6 H_{2} O \to 4 H_{3} P O_{4}$

In the firest step, for 75% yield, moles of $P_{4} O_{10}$ obtained.
= moles of $P_{4} \times \frac{75}{100}$
$= \frac{310}{124} \times \frac{75}{100}$
$= 1.875 m o l e s$

Again,
$P_{4} O_{10} + 6 H_{2} O \to 4 H_{3} P O_{4}$
Moles of water given $= \frac{54}{18} = 3$ moles
Actually amount of water required for 1.875 moles of $P_{4} O_{10}$ is
$= 1.875 \times 6$
$= 11.25$ moles of water.
$∴$ Water is limiting agent.
$∴$ 6 moles of $H_{2} O$ produced 4 moles of $H_{3} P O_{4}$
3 moles of $H_{2} O$ will produce
$= \frac{4}{6} \times 3$ moles of $H_{3} P O_{4}$
$= 2$ moles of $H_{3} P O_{4}$
Molar mass of $H_{3} P O_{4} = 98 g$
Therefore, amount of $H_{3} P O_{4}$ obtained = $2 \times 98 g = 196 g$

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Similar questions

Q. In a particular experiment,

310 g

of phosphorous (

P_{4}

) reacted with excess of oxygen to form

P_{4} O_{10}

in 75% yield. Then

P_{4} O_{10}

formed reacted with

54

g of water to give

H_{3} P O_{4} .

Amount of

H_{3} P O_{4}

obtained (in gm) is :
(Consider only two places after decimal for calculation)

Q. Phosphoric acid (

H_{3} P O_{4}

) is prepared in a two-step process.

P_{4} + 5 O_{2} \to P_{4} O_{10}

P_{4} O_{10} + 6 H_{2} O \to 4 H_{3} P O_{4}

186 g of phosphorus is made to react with excess oxygen which forms

P_{4} O_{10}

that has a 60 % yield. In the second reaction, 80% yield of

H_{3} P O_{4}

is obtained. The mass of

H_{3} P O_{4}

produced is:
(Molar mass of phosphorous = 31 g/mol)

Q. Phosphoric acid

(H_{3} P O_{4})

is prepared in a two-step process:
(1)

P_{4} + 5 O_{2} \to P_{4} O_{10}

(2)

P_{4} O_{10} + 6 H_{2} O \to 4 H_{3} P O_{4}

In step (1),

62 g

of phosphorus is made to react with excess oxygen to form

P_{4} O_{10}

with 85 % yield. In step (2), the yield of

H_{3} P O_{4}

obtained is 90 %. Calculate the mass of

H_{3} P O_{4}

produced.

Q. Phosphoric acid

(H_{3} {PO}_{4})

prepared in a two step process.
(1)

P_{4} {+ 5O}_{2} \to P_{4} O_{10}

(2)

P_{4} O_{10} + {6H}_{2} O \to {4H}_{3} {PO}_{4}

We allow 62 g of phosphorus to react with excess oxygen which form

P_{4} O_{10}

which gives 85% yield. If in the step (2) reaction, 90% yield of

H_{3} {PO}_{4}

is obtained then mass of

H_{3} {PO}_{4}

produced is:

Q. Phosphoric acid

(H_{3} P O_{4})

is prepared in a two-step process:
(1)

P_{4} + 5 O_{2} \to P_{4} O_{10}

(2)

P_{4} O_{10} + 6 H_{2} O \to 4 H_{3} P O_{4}

In step (1),

62 g

of phosphorus is made to react with excess oxygen to form

P_{4} O_{10}

with 85 % yield. In step (2), the yield of

H_{3} P O_{4}

obtained is 90 %. Calculate the mass of

H_{3} P O_{4}

produced.

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In a particular experiment, 310 g of phosphorous (P4) reacted with excess of oxygen to form P4O10 in 75% yield. Then P4O10 formed reacted with 54 g of water to give H3PO4. Amount of H3PO4 obtained (in gm) is : (Consider only two places after decimal for calculation)