# How much oxygen (in grams) will from 25.7g of N2O5

### Question:

Q1 Answer :  reaction given the question is :

2 N2O5 (s) => 4 NO2 (g) + O2 (g)

Also the given mass of N2O5 = 25.7 gram

Now from the above balanced chemical reaction, we can say that   216 gram of N2O5 decompose to give 184 gram of NO2 and 32 gram of O2.

Therefore we can say that

Amount of O2 formed by 216 gram of N2O5 = 32 gram

Therefore, Amount of O2 formed by 25.7 gram of N2O5 = (32/216)×25.7 = 3.8074 gram

Note : In the above solution , Molar mass of N2O5 i.e. 108 gram & O2 i.e. 32 gram has been used.

Q2 Answer : The balanced chemical reaction of the question is :

Pb(NO3)2 (aq) + 2 KI (aq) => 2 KNO3 (aq) + PbI2 (s)↓

Also given that,

Molar concentration of Pb(NO3)2 (aq) = 0.0509 M

Volume of Pb(NO3)2 (aq) = 63.8 ml

Molar concentration of KI (aq) = 0.0436 M

Volume of KI (aq) = 85.2 ml

Therefore with help of Molar concentration we can say that:

Mass of Pb(NO3)2 in 1000 ml solution = 0.0509mol × molar mass of Pb(NO3)2 = 0.0509×331.2 = 16.858 gram

Similarly, Mass of Pb(NO3)2 in 63.8 ml solution = (16.858/1000)×63.8 = 1.0755 gram —– (i)

In a similar way, Mass of KI in 1000 ml solution = 0.0436mol × molar mass of KI = 0.0436×166 = 7.2376 g

Therefore, Mass of KI in 85.2 ml solution = (7.2376/1000)×85.2 = 0.6166 gram —— (ii)

Now from the balanced chemical reaction we can say that

331.2 gram of Pb(NO3)2 react with 332 gram of KI

gram of Pb(NO3)2 react with (332/331.2) gram KI

Therefore, amount of KI which will react with 1.0755gram of Pb(NO3)2 will be = (332/331.2)×1.0755 = 1.078 gram of KI

But, from equation —- (ii), the amount of KI present is 0.6166gram only.

Therefore, we can compute here that in this reaction KI will finish first. And KI is the limiting reagent.

Therefore, the amount of precipitation will depend upon the amount of limiting reagent of the reaction.

Now, from the balanced chemical reaction :

Amount of PbI2 (s) produced from 332 gram of KI = 461.01 gram

Therefore, amount of PbI2 (s) produced from 0.6166 gram of KI = (461.01/332)×0.6166 = 0.8562 gram

Therefore, the required answer is 0.8563 gram.

Q3 Answer : Given reaction of the question is :

4 Al (s) + 3 O2 (g) => 2 Al2O3 (s)

Also the given mass of Al = 4.34 gram

Now from the above balanced chemical reaction we can say :

108 gram of Al react with 96 gram of O2

Therefore, amount of O2 reacting with 108 gram Al = 96 gram

Amount of O2 reacting with 4.34 gram Al = (96/108)×4.34

= 3.8578 gram

Therefore, number of moles of O2 reacting with 4.34 gram Al = (3.8578 / molar mass of O2)  = (3.8578/32) = 0.12055 mol.

Now the Volume occupied by 0.12055 mol O2 can obtained from ideal gas equation as,

PV = nRT

Given in the question, Pressure = 2.99 atm

No. of moles = 0.12055 mol

Gas constant , R = 0.082057 (L.atm/mol.K)

Temperature, T = 608 K