Q-7)
The balanced equation of the given reaction
#Fe_2O_3+3CO->2Fe+3CO_2#
Molar mass of #Fe_2O_3=(2xx56+3xx16)=160g"/mol"#
Reacting mass of #Fe_2O_3=1kg=1000g=(1000g)/(160g"/mol")=6.25mol#
Molar mass of #CO=12+16=28g"/mol"#
Let reacting mass of #CO=xg=(xg)/(28g"/mol")=x/28mol#
As per balanced equation the reacting mole ratio of
#Fe_2O_3:CO=1:3#
So #6.25/(x/28)=1/3#
#=>x=3xx6.25xx28g=525g=0.625kg#
Q-8)
The balanced equation of the given reaction
#2Fe+3Cl_2->2FeCl_3#
Molar mass of #Fe=56g"/mol"#
Reacting mass of #Fe=20g=(20g)/(56g"/mol")=5/14mol#
Molar mass of #Cl_2=2xx35.5=71g"/mol"#
Let reacting mass of #CO=yg=(yg)/(71g"/mol")=y/71mol#
As per balanced equation the reacting mole ratio of
#Fe:Cl_2=2:3#
So #(5/14 )/(y/71)=2/3#
#=>y=(3xx5xx71)/(2xx14)g=38.03g#
Q-9)
The balanced equation of the given reaction
#2NH_3+H_2O_2->N_2H_4+H_2O#
Molar mass of #NH_3=14+3xx1=17g"/mol"#
Let the required reacting mass of
#NH_3=zg=(zg)/(17g"/mol")=z/17mol#
Molar mass of #N_2H_4=2xx14+4xx1=32g"/mol"#
The produced mass of #N_2H_4=100g=(100g)/(32g"/mol")=25/8mol#
As per balanced equation the mole ratio of
#NH_3:N_2H_4=2:1#
So #(z/17 )/(25/8)=2/1#
#=>z=(2xx25xx17)/(8)g=106.25g#
Q-10)
The balanced equation of the physical change for the dehydration of hydrated #Na_2SO_4.xH_2O# is given as follows
#Na_2SO_4.xH_2O->Na_2SO_4+xH_2O#
Formula mass of #Na_2SO_4.xH_2O#
#=(2xx23+32+4xx16+x(2xx1+16))g"/mol"#
#=(142+18x)g"/mol"#
Formula mass of anhydrous #Na_2SO_4#
#=(2xx23+32+4xx16)g"/mol"#
#=142g"/mol"#
It is obvious from the equation that
#(142+18x)/142=10/4.4=25/11#
#=>18x=(25xx142)/11-142=(14xx142)/11#
#=>x=(14xx142)/(11xx18)~~10#
So Formula mass of #Na_2SO_4. 10H_2O#
#=(2xx23+32+4xx16+10xx(2xx1+16))g"/mol"#
#=322g"/mol"#
