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Calculate the mas of Magnesium required to completely react with 250cm^(3) of 0.1 MHCl. |
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Answer» SOLUTION :`250xx0.1M=25` millimoles of HCl From the EQUATION it is clear that 2HCl is reacting with ONE Mg. `:.` 25 millimoles of HCl react with 12.5 millimoles of Mg `:.` Mass of magnesium required `=` volume of Mg `xx` ATOMIC weight `=12.5xx10^(3)xx24=0.3g` of Mg OR `V_(HCl)=100cm^(3), V_(NaOH)=25cm^(3) , M_(HCl)=?, M_(NaOH)=0.1M` and `V_(HCl)xxM_(HCl)=V_(NaOH)xxM_(NaOH)xxa_(NaOH)` Where `a_(HCl)=` basicity of HCl `=1, aNaOH=` acidity of `NaOH=1` `M_(HCl)=(25xx0.1)/100=0.025`m Mass of HCl PRESENT in `100cm3=(M_(HCl)xx"Molecular wieight of " HCl)/10` `=(0.025xx36.5)/10=0.0912g` |
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