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Answer 1 · 2017-01-17T16:21:03

Let the mass of #Mg# in #36g# mixture be #xg# and the mass of Ca be #(36-x)g#
Considering atomic masses as follows

#Mg->24g"/mol"#

#Ca->40g"/mol"#

#Br->80g"/mol"#

Balanced equations of ractions are
#Mg+Br_2=MgBr_2......[1]#

#Ca+Br_2=CaBr_2........[2]#

As per equation [1]

#24g# Mg forms #184g# #MgBr_2#

#xg" "# Mg forms #(184x)/24g " "MgBr_2#

As per equation [2]

#40g# Ca forms #200g# #CaBr_2#

#(36-x)g" "# Ca forms #((36-x)200)/20g " "CaBr_2#

Hence by the given condition of the problem we can write

#184/24x+(36-x)200/40=212#

#=>23/3x+(180-5x)=212#

#=>23/3x-5x=212-180=32#

#=>(23x-15x)/3=32#

#=>(8x)/3=32#

#=>x=12#

So the mass of Mg is #12g#

and the mass of Ca is #(36-12)g=24g#

#% "of Mg"=12/36xx100%=33 1/3%#

#% "of Ca"=24/36xx100%=66 2/3%#