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`        For the formation of 3.65 gram of HCL gas, what volume of hydrogen gas and chlorine gas are required at NTP condition ? (1) 1L,1L (2) 1.12L , 2.24L (3) 3.65 L, 1.83L (4) 1.12L, 1.12L`
5 months ago

Saurabh Koranglekar
3296 Points
```							Dear studentAt mass of HCL is 36.5And reaction is H2 + Cl2 = 2 HCL0.1 mole of HCl req 0.05 moles of reactants0.05 *22.4 = 1.12 litRegards
```
5 months ago
Vikas TU
9801 Points
```							Moles of HCl = given mass / molar mass  = 3.65 /36.5 = 0.1 mole 0.5H2 + 0.5 Cl2 = HCl From the given balanced equation:1 mole of HCl require = 0.5 mole of Hydrogen and 0.5  mole of chlorine0.01 mole of HCl require=0.05 mole of Hydrogen and 0.05 mole of chlorine.According to Avogadro's law, 1 mole of gas occupies=22.4 L of gas at STP.Thus 0.05 mole of every gas occupies= Litres at STP.
```
5 months ago
Vikas TU
9801 Points
```							Dear student the ans would be 1.12 Litre at STP , Hope you can solve and understand the method ,Good luck cheers
```
5 months ago
16 Points
```							As the Ratio of H to Cl in HCl is 1:1we know that at a constant temperature the volume of equal moles of two gases is equalVolume of H2 gas = volume of Cl2 gasThis is true only for option A and D Therefore options C and D can be neglected  By avagadro ,molar mass of 1 mole of HCl = molar mass of H + molar mass = 1+35=36gWe know that 36g or 1mole of HCl occupies 1 liter volume at STP => 3.68 grams would hold 3.68/36 = 1.24lTherefore we would require equal amounts of H and Cl to make HCl Volume of H = Volume of Cl = Volume of HCl /2 = 0.1 /2 = 0.05l
```
5 months ago
16 Points
```							This is the continuation of the before and Volume of H and Cl = 1.24/2 l = 0.62 l We want the volume of H2 gas We know that the volume of a diatomic gas is half of that of its monoatomic gas => volume H2 gas and Cl2 gas which is required to make 3.68g of HCl = 0.62 /2 = 0.31L  Thus the correct amns is none of the above (which is not an option) Therefore you van conclude that the question is incorrect
```
5 months ago
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