Question

Zinc reacts with hydrochloric acid according to the reaction equation Zn ( s ) + 2 HCl ( aq ) ⟶ ZnCl 2 ( aq ) + H 2 ( g ) How many milliliters of 5.50 M HCl (aq) are required to react with 6.25 g Zn (s) ?

Answer 1

Answer:

34.7mL

Explanation:

First we have to convert our grams of Zinc to moles of zinc so we can relate that number to our chemical equation.

So: 6.25g Zn x (1 mol / 65.39 g) = 0.0956 mol Zn

All that was done above was multiplying the grams of zinc by the reciprocal of zincs molar mass so our units would cancel and leave us with moles of zinc.

So now we need to go to HCl!

To do that we multiply by the molar coefficients in the chemical equation:

$\frac{0.0956g Zn}{1 } (\frac{2 mol HCl}{1molZn})$

This leaves us with 2(0.0956) = 0.1912 mol HCl

Now we use the relationship M= moles / volume , to calculate our volume

Rearranging we get that V = moles / M

Now we plug in: V = 0.1912 mol HCl / 5.50 M HCl

V= 0.0347 L

To change this to milliliters we multiply by 1000 so:

34.7 mL