Answer to Question #22098 in Inorganic Chemistry for khadijah
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
2Fe + 3Cl2 = 2FeCl3
n(Fe) = (4.5mol*3)/2 = 6.75 mol
n(Fe) = m/M
n(Fe) = 240g/56g/mol = 4.285 mol
n(FeCl3) = 4.285 mol
M(FeCl3) = 56 + 35.5*3 = 162.5 g/mol
m(FeCl3) = 162.5g/mol * 4.285 mol = 696 g
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