Question

P+5HNO3(conc)→H3PO4+H20+5NO2.....if 9.3 g of phosphorus was used in the reaction :

a)Number of moles of phosphorus taken

b)the mass of phosphoric acid formed

c)the volume of NO2 produced at stp [H=1,N=14,P=31,O=16]

Answer 1

Answer:

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Explanation:

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Answer 2

Number of moles of phosphorus taken is 0.3 mole

The mass of phosphoric acid formed is 29.4 g

The volume of $NO_{2}$ produced at STP is 33.6 L

Given : The chemical reaction $P +5HNO_{3} (conc) - > H_{3}PO_{4} +H_{2}O +5NO_{2}$

9.3 g of phosphorus was used in the reaction

To Find : a)Number of moles of phosphorus taken

b)the mass of phosphoric acid formed

c)the volume of NO2 produced at STP  [H=1,N=14,P=31,O=16]

Solution :

The following chemical reaction is given : $P +5HNO_{3} (conc) - > H_{3}PO_{4} +H_{2}O +5NO_{2}$

and 9.3 g of phosphorus was used in the reaction

We have to find

a) Number of moles of phosphorus taken  

Now we know that

Number of moles =  Given Mass/ Molar Mass

Here Given mass of phosphorus is 9.3 g

Molar mass of phosphorus is 31

So molar mass is $\frac{9.3}{31}$

= 0.3 mole

So number of moles of phosphorus taken is 0.3 mole

b) the mass of phosphoric acid formed

Now in the reaction 1 mole of phosphorus forms 1 mole of phosphoric acid and 5 mole of $NO_{2}$

So 0.3 mole of phosphorus produce 0.3 mole of of phosphoric acid and 1.5 mole of $NO_{2}$

Now

Number of moles =  Given Mass/ Molar Mass

Given mass = Number of moles × Molar mass

Number of moles of phosphoric acid is 0.3

Molar mass of phosphoric acid is 3+31+64 = 98

Mass of phosphoric acid is

0.3 × 98

= 29.4 g

So the mass of phosphoric acid formed is 29.4 g

c) the volume of $NO_{2}$ produced at STP

We know that

Number of moles = V/22.4

V is volume

Volume = Number of moles × 22.4

Number of moles of $NO_{2}$  is 1.5

Volume = 1.5 × 22.4

= 33.6 L

So the volume of $NO_{2}$ produced at STP is 33.6 L

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