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Answer 1 · 2016-06-07T13:37:25

$x =3$

The reaction is:

$MgCO_3*xH_2OtoMgCO_3+xH_2O$

The masses involved can be identified:

$1.25g to 0.759g+0.491g$ (the mass of water being the balancing figure from the information given)

If we determine the relative formula mass for the various components in the above reaction, we can determine the number of moles involved:

$MgCO_3=24.305+12.011+(15.999times3)=84.313$
$H_2O=(1.008*2)+15.999=18.015$

Hence the number of moles on the right hand side of the equation can be determined:
$MgCO_3=0.759/84.313=9.00times10^-3"moles"$

$H_2O=0.491/18.015=0.02725 "moles"$

The ratio of $H_2O$ to $MgCO_3$ = $0.02725/(9.00times 10^-3)=3.028$
Hence $x= 3$