Reacting Zn with .248M HCl to Produce H2

[kippy]

How many milliters of .248 M HCl are needed to react with 1.36g of zinc to produce hydrogen gas

 

[Gokul43201]

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[ravenprp]

To determine the amount of HCl needed to react with 1.36g of zinc, we can use the balanced chemical equation for the reaction:

Zn + 2HCl → ZnCl2 + H2

From the equation, we can see that 1 mole of zinc reacts with 2 moles of HCl to produce 1 mole of hydrogen gas. Therefore, we need to calculate the moles of zinc in 1.36g.

1.36g of Zn x 1 mol Zn/65.38g Zn = 0.0208 mol Zn

Since the ratio of zinc to HCl is 1:2, we need twice the amount of HCl in moles.

0.0208 mol Zn x 2 mol HCl/1 mol Zn = 0.0416 mol HCl

Now, we can use the molarity formula to calculate the volume of 0.248 M HCl needed.

Molarity = moles of solute/ volume of solution

0.248 M = 0.0416 mol HCl/ volume of solution

Volume of solution = 0.0416 mol HCl/0.248 M = 0.1678 L

Since 1 L = 1000 mL, the volume of 0.248 M HCl needed is 167.8 mL. Therefore, 167.8 mL of 0.248 M HCl is needed to react with 1.36g of zinc and produce hydrogen gas.