(06.04 HC)

Part 1. A chemist reacted 15.0 liters of F2 gas with NaCl in the laboratory to form Cl2 and NaF. Use the ideal gas law equation to determine the mass of NaCl that reacted with F2 at 280. K and 1.50 atm.

F2 + 2NaCl → Cl2 + 2NaF

Part 2. Explain how you would determine the mass of sodium chloride that can react with the same volume of fluorine gas at STP

can someone explain this to me

Respuesta :

The mass of sodium chloride at the two parts are mathematically given as

  • m=10,688.18g
  • mass of Nacl(m)=39.15g

What is the mass of sodium chloride that can react with the same volume of fluorine gas at STP?

Generally, the equation for ideal gas is mathematically given as

PV=nRT

Where the chemical equation is

F2 + 2NaCl → Cl2 + 2NaF

Therefore

1.50x15=m/M *(1.50*0.0821)

1-50 x 15=m/58.5 *(1.50*0.0821)

m=10,688.18g

Part 2

PV=m'/MRT

1*15=m'/58.5*0.0821*273

m'=39.15g

mass of Nacl(m)=m'=39.15g

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