Answer:
5 g
Explanation:
From the question, we have,
Molarity of FeIII solution= 0.245 M
Volume of solution = 125 ml
From
number of moles= concentration × volume
We have;
Number of moles= 0.245 M × 125/1000
Number of moles = 0.031 moles
Molar mass of Fe III = 162.5g/moles
Mass of iron III = number of moles× molar mass = 0.031 × 162.5= 5 g
n
The mass of sodium chloride at the two parts are mathematically given as
- m=10,688.18g
- mass of Nacl(m)=39.15g
<h3>What is the mass of sodium chloride that can react with the same volume of fluorine gas at STP?</h3>
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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