Question

Zinc reacts with hydrochloric acid according to the reaction equation

Zn(s) + 2 HCl(aq) → ZnCl₂(aq) + H₂(g)
How many milliliters of 1.50 M HCl(aq) are required to react with 5.05 g Zn(s)?
volume: ___ mL

Answer 1

First, we need to calculate the number of moles of Zn:

m(Zn) = 5.05 g

M(Zn) = 65.38 g/mol (molar mass of Zn)

n(Zn) = m(Zn) / M(Zn) = 5.05 g / 65.38 g/mol = 0.0773 mol

According to the balanced chemical equation, 1 mole of Zn reacts with 2 moles of HCl. Therefore, the number of moles of HCl needed is:

n(HCl) = 2 × n(Zn) = 2 × 0.0773 mol = 0.1546 mol

Now we can use the molarity of the HCl solution to calculate the volume needed:

M(HCl) = 1.50 mol/L

n(HCl) = V(HCl) × M(HCl)

V(HCl) = n(HCl) / M(HCl) = 0.1546 mol / 1.50 mol/L = 0.103 L = 103 mL

Therefore, we need 103 mL of 1.50 M HCl(aq) to react with 5.05 g Zn(s).

Answer 2

many milliliters of 1.50 M HCl(aq) are required to react with 5.05 g Zn(s)?

volume: 16 mL