How many milliliters of 3.00 M #HCI(aq)# are required to react with 8.55 g of #Zn(s)#?

1 Answer

Answer:

#V_"HCl"=0.087L#

Explanation:

Molecular mass HCl=36.4g , Zn=65.4g

#2HCl + Zn -> ZnCl_2 + H_2#

n°mol=#"8.55g"/"65.4g"# = 0.13mol

In the reaction, we can see HCl doubles the n°mol of Zn, so...
#2(0.13)= 0.26mol#

#M="n°mol"/"Volume"#

#3="0.26mol"/V#

#V=0.087L# of 3M HCl