Final answer:
The mass of calcium that was dropped into the hydrochloric acid, denoted as x, is approximately 0.0842 grams. This was calculated using titration formulas and stoichiometry based on the neutralization reaction between calcium and hydrochloric acid.
Step-by-step explanation:
To find the value of x, which is the mass of calcium dropped into hydrochloric acid, we need to set up a stoichiometric calculation based on the neutralization reactions that have occurred. In this scenario, we know that calcium reacts with hydrochloric acid to form calcium chloride and hydrogen gas. After the reaction, the remaining HCl was used to titrate NaOH. The volume and concentration of NaOH and the volume of HCl required for neutralization allow us to calculate the remaining moles of HCl, which in turn gives us the initial moles of HCl that reacted with calcium.
We have 14 cm³ of the resulting solution neutralizing 10 cm³ of a 0.03 mol/dm³ NaOH solution. Using the titration formula M1V1 = M2V2, where M represents molarity and V represents volume, we can calculate the moles of HCl neutralized by NaOH. Therefore, (0.10 mol/dm³) * (0.014 dm³) = (0.03 mol/dm³) * (0.010 dm³); hence, x = 0.0042 moles of HCl. Considering the stoichiometry of the reaction between HCl and calcium, we know that one mole of calcium reacts with two moles of HCl. This implies that x/2 = moles of calcium. To find the mass of calcium, we simply multiply the moles of calcium by its molar mass (40.08 g/mol).
Therefore, the mass x of calcium is (0.0042/2) * 40.08 g/mol, which equals approximately 0.0842 grams.