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

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Answer 1 · 2017-06-26T01:43:18

$V_"HCl"=0.087L$

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

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