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    Can anyone please explain this thing to me?? This is the question that is asked:
    1) A student reacts 9.2g of sodium with 0.8 g of sulfur to form sodium sulfide,Na2S. Which reactant is in excess? Relative atomic values - Na=23, S=32
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    (Original post by Angelin Ravi)
    Can anyone please explain this thing to me?? This is the question that is asked:
    1) A student reacts 9.2g of sodium with 0.8 g of sulfur to form sodium sulfide,Na2S. Which reactant is in excess? Relative atomic values - Na=23, S=32
    You want that someone explains you the steps which were used to get the solution?

    *Na = 23 and S = 32 in the first step stand for the atomic masses. In the first step, the amount of substance was calculated. This one depends on the mass of the substance and the mass of the atoms (molecular mass) *

    In the second step, the ratio of moles are written down. Na has 2 moles and S has 1 mol. From the view of S, the ratio is 1:2.

    In the third and last step, it was calculated how much mol of S (molecular mass) has to react with Na to get completed. In this calculation, the ratio (1:2) was multiplied by the amount of substance of Na (o,4 mol). After that this, the result (0,2 mol) was subtracted to the amount of S from the first step (0,25 mol) to find out how much excess of S exists. It is 0,05 mol.

    * In other words: there is more amount of substance of S in existance than really needed. Understand?*
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    (Original post by Kallisto)
    You want that someone explains you the steps which were used to get the solution?

    *Na = 23 and S = 32 in the first step stand for the atomic masses. In the first step, the amount of substance was calculated. This one depends on the mass of the substance and the mass of the atoms (molecular mass) *

    In the second step, the ratio of moles are written down. Na has 2 moles and S has 1 mol. From the view of S, the ratio is 1:2.

    In the third and last step, it was calculated how much mol of S (molecular mass) has to react with Na to get completed. In this calculation, the ratio (1:2) was multiplied by the amount of substance of Na (o,4 mol). After that this, the result (0,2 mol) was subtracted to the amount of S from the first step (0,25 mol) to find out how much excess of S exists. It is 0,05 mol.

    * In other words: there is more amount of substance of S in existance than really needed. Understand?*
    Thanks,but, why would you find how much of S has to react with Na? Why can't this be done the other way round?
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    (Original post by Angelin Ravi)
    Thanks,but, why would you find how much of S has to react with Na? Why can't this be done the other way round?
    It can't be done in the other way round, because you need to know the molecular masses and the ratio of them. That is why the first two steps are neccessary. The third step is a calculation of those results.*

    Without the ratios and the molecular masses, this kind of task cannot be solved. And I don't really not an alternatice way to get it out. *
 
 
 
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