How many milliliters of 5.50 M HCl(aq) are required to react with 3.05 g Zn(s)?

Answer 1

we need to first write a balanced chemical equation for the reaction between zinc and hydrochloric acid:

Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)

From the balanced equation, we see that 1 mole of Zn reacts with 2 moles of HCl. To find the number of moles of Zn, we can use its molar mass:

molar mass of Zn = 65.38 g/mol

moles of Zn = mass of Zn / molar mass of Zn

moles of Zn = 3.05 g / 65.38 g/mol

moles of Zn = 0.0467 mol

Since 1 mole of Zn reacts with 2 moles of HCl, we need twice as many moles of HCl to react completely with the given amount of Zn:

moles of HCl = 2 × moles of Zn

moles of HCl = 2 × 0.0467 mol

moles of HCl = 0.0934 mol

Now we can use the molarity of the HCl solution and the number of moles of HCl to calculate the volume of the solution required:

moles of HCl = molarity × volume (in liters)

volume (in liters) = moles of HCl / molarity

Substituting the values we obtained:

volume (in liters) = 0.0934 mol / 5.50 mol/L

volume (in liters) = 0.01698 L

Finally, we can convert the volume from liters to milliliters:

volume (in milliliters) = 0.01698 L × 1000 mL/L

volume (in milliliters) = 16.98 mL

Therefore, we need approximately 16.98 mL of 5.50 M HCl(aq) to react with 3.05 g Zn(s).