How many milliliters of 4.00 M $H C l (a q)$ $HCl(aq)$ are required to react with 8.75 g of $Z n (s)$ $Zn(s)$ ?

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Answer 1 · 2017-06-20T02:54:56

Approx. $70*mL..........$

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.

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

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

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