2F + 3Cl2 -> 2FeCl3
a. How many grams of Fe are needed to combine with 4.5 moles of Cl2?
b. If 240 g of Fe is to be used in this reaction, with adequate Cl2, how many moles of FeCl3 will be produced?
a) V(Fe)=2/3*V(Cl2)=4.5mol*2/3=3mol
M(Fe)=56g/mol
m(Fe)=3mol*56g/mol=168g
b) V(Fe)=240g/56g/mol=4.29mol
V(Fe)=V(FeCl3)=4.29mol
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