How many milliliters of 4.00 M #HCl(aq)# are required to react with 8.75 g of #Zn(s)#?

1 Answer
Jun 20, 2017

Answer:

Approx. #70*mL..........#

Explanation:

We need (i) a stoichiometrically balanced equation......

#Zn(s) + 2HCl(aq) rarr ZnCl_2(aq) + H_2(g)uarr#

And (ii) equivalent quantities of metal and hydrochloric acid.

#"Moles of zinc"=(8.75*g)/(65.39*g*mol^-1)=0.134*mol.#

#"Moles of HCl"=(0.134*molxx2)/(4.00*mol*L^-1)xx10^3*mL*L^-1=67*mL.#