Question

Calcium metal reacts with hydrochloric acid (HCl) to form aq calcium chloride and hydrogen gas.

Ca(s)+2HCl(aq) -----------> CaCl2(aq)+H2(g)
0.401 g of calcium metal is reacted with 0.350L of 0.0400M hydrochloric acid.
(a) Write the complete ionic equation of the reaction.
(b) Calculate the moles of calcium and hydrochloric acid.
(c) Identify the limiting reagent.
(d) Assuming complete reation, determine the moels of excess reagent that reamains in the reaction.

Answer 1

a. Ca(s) + 2H⁺ → Ca²⁺ + H₂(g)

b. 0.0100 moles Ca; 0.0140 moles HCl.

c. HCl is limiting reactant.

d. 0.0030 moles of Ca

Step-by-step explanation:

a. In water, HCl dissociates in H⁺ and Cl⁻ and CaCl₂ in Ca²⁺ and 2 Cl⁻. The ionic equation is:

Ca(s) + 2H⁺ + 2Cl⁻ → Ca²⁺ + 2Cl⁻ + H₂(g)

Ca(s) + 2H⁺ → Ca²⁺ + H₂(g)

b. Moles Ca -Molar mass 40.1g/mol-:

0.401g * (1mol / 40.1g) = 0.0100 moles Ca

Moles HCl:

0.350L * (0.0400mol / L) = 0.0140 moles HCl

c. In the reaction, 1 mole of Calcium reacts with 2 moles of HCl.

For a complete reaction of 0.0100 moles of Calcium there are necessaries:

0.0100mol Ca * (2 mol HCl / 1 mol Ca) = 0.0200 moles HCl

As there are just 0.0140 moles of HCl,

HCl is limiting reactant

d. For a complete reaction of 0.0140 moles of HCl, the moles of Ca that react are:

0.0140 moles HCl * (1 mol Ca / 2 mol HCl) = 0.0070 moles Ca

That means will remain:

0.0100 moles - 0.0070 moles:

0.0030 moles of Ca