Question #74d98

1 Answer
P dilip_k
Oct 7, 2016

The HCl solution taken 20mL 0.1M
The no. Of moles of HCl in this solution
#="volume in L"xx "molarity"=20xx10^(-3)xx0.1=2*10^-3" moles"#

The NaOH solution required for neutralisation of excess acid is 7.5mL 0.2M
The no. Of moles of NaOH in this solution
#="volume in L"xx "molarity"#
#=7.5xx10^(-3)xx0.2=1.5*10^-3" moles"#

The equation of the acid base reaction is

#HCl+NaOH->NaCl+H_2O#

This equation reveals that acid and base neutralises in 1:1 mole ratio.

So #(2*10^-3-1.5*10^-3)=0.5*10^-3#moles HCl reacts with Mg

Now equation of the reaction of Mg with HCl is

#2HCl+Mg->MgCl_2+H_2#

This equation reveals that HCl and Mg reacts completely with each other in mole ratio 2:1.

So

# 0.5*10^-3" mol " HCl# will be neutralised by #1/2*0.5*10^-3" mol "Mg#

Hence the mass of Mg added to HCl solution was

#=1/2*0.5*10^-3" mol"xx24 g/"mol"#

#=6*10^-3g " "Mg#

where atomic mass of Mg #=24g/"mol"#

Related questions
See all questions in Neutralization
Impact of this question
1418 views around the world
You can reuse this answer
Creative Commons License