How many milliliters of 3.00 M $H C I (a q)$ $HCI(aq)$ are required to react with 8.55 g of $Z n (s)$ $Zn(s)$ ?

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Answer 1 · 2017-06-26T01:43:18

$V_{HCl} = 0.087 L$ $V_"HCl"=0.087L$

Molecular mass HCl=36.4g , Zn=65.4g

$2 H C l + Z n \to Z n C l_{2} + H_{2}$ $2HCl + Zn -> ZnCl_2 + H_2$

n°mol= $\frac{8.55g}{65.4g}$ $"8.55g"/"65.4g"$ = 0.13mol

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

$M = \frac{n°mol}{Volume}$ $M="n°mol"/"Volume"$

$3 = \frac{0.26mol}{V}$ $3="0.26mol"/V$

$V = 0.087 L$ $V=0.087L$ 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)?