Question

(06.04 HC)

Answer 1

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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