Question

From the equation $ 16Fe+{{S}_{8}}\to 8F{{e}_{2}}{{S}_{3}}. $ How much $ F{{e}_{2}}{{S}_{3}} $ would be produced in grams if you started with $ 7g $ of $ Fe $ ?

Answer 1

Hint: Here we have iron which reacts with sulphur to give Iron sulphate $ F{{e}_{2}}{{S}_{3}} $ so for finding out the amount in grams of $ F{{e}_{2}}{{S}_{3}} $ we must know the molar mass of each constituents forming $ F{{e}_{2}}{{S}_{3}} $ that is $ Fe $ and $ S $ . We can easily find out the amount with the stoichiometry as given in the equation so firstly find out the mass of $ F{{e}_{2}}{{S}_{3}} $ and then see how many moles are forming according to the equation.

Complete answer:
When a molecule reacts with another molecule it will form a compound having just different t physical and chemical properties. For example if we consider water it is prepared by reaction of hydrogen gas and oxygen gas in presence of electric arc.
Now see both reactants are gases but the product is a liquid. So, in this way every reaction forms some product which has its individual chemical and physical properties. While writing the equation we should note that the number of atoms on the left hand and right hand side should be equal. It means that the equation should be balanced. Now according to the above equation we have iron which reacts with sulphur and forms Iron sulphate. The following chemical reaction is given by;
 $ 16Fe+{{S}_{8}}\to 8F{{e}_{2}}{{S}_{3}} $
So, from the above equation we can say, $ 16 $ moles of $ Fe $ after reacting with $ 1 $ mole of $ {{S}_{8}} $ will give, $ 8 $ moles of $ F{{e}_{2}}{{S}_{3}} $ , now, molecular weight of $ Fe $ is $ 56gmo{{l}^{-1}} $ and molecular weight of $ F{{e}_{2}}{{S}_{3}} $ is $ (56.2+32.3)gmo{{l}^{-1}}=208gmo{{l}^{-1}} $
So, we can say, due to this reaction, we will get, $ 208.8g~ $ of $ F{{e}_{2}}{{S}_{3}} $ from $ 56.16g~ $ of $ Fe $ so, if we perform the reaction with $ 7g $ of $ Fe $ we will get; $ \dfrac{208.8}{56.16}\times 7g=13g(F{{e}_{2}}{{S}_{3}}) $

Note:
The number of moles of $ F{{e}_{2}}{{S}_{3}} $ are $ 8 $ moles of $ F{{e}_{2}}{{S}_{3}} $ thus we multiply the mass of $ F{{e}_{2}}{{S}_{3}} $ with $ 8. $ Thus we can easily but the amount of any of the reactants and products in any reaction by multiplying the molar mass with the stoichiometric coefficient. The important point about the reaction is conversion of $ 1amu $ into grams. If we multiply the mass of any element.