Answer:
A. 16.0 cm³ is correct
Explanation:
To determine the volume of hydrochloric acid required to react with calcium carbonate, we need to use the balanced chemical equation and the given information.
The balanced chemical equation for the reaction between hydrochloric acid (HCl) and calcium carbonate (CaCO3) is:
CaCO3 + 2HCl → CaCl2 + H2O + CO2
From the equation, we can see that 1 mole of calcium carbonate reacts with 2 moles of hydrochloric acid to produce 1 mole of carbon dioxide gas.
Given that the volume of carbon dioxide gas produced is 227 cm³ at STP (standard temperature and pressure), we need to find the corresponding number of moles of carbon dioxide using the ideal gas law.
Using the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature, we can rearrange the equation to solve for the number of moles:
n = PV/RT
At STP, the pressure (P) is 1 atm and the temperature (T) is 273 K. The ideal gas constant (R) is 0.0821 L·atm/(mol·K).
Converting the volume of carbon dioxide from cm³ to liters (L):
227 cm³ = 227/1000 L = 0.227 L
Using these values in the equation, we can calculate the number of moles of carbon dioxide:
n = (1 atm) * (0.227 L) / (0.0821 L·atm/(mol·K) * 273 K) = 0.0106 mol
Since the balanced equation shows that 1 mole of calcium carbonate reacts with 1 mole of carbon dioxide, we can conclude that 0.0106 mol of calcium carbonate reacted.
As stated in the question, calcium carbonate is in excess. Therefore, the amount of hydrochloric acid needed will be twice the number of moles of carbon dioxide produced:
2 * 0.0106 mol = 0.0212 mol
To find the volume of hydrochloric acid needed, we can use the given concentration of hydrochloric acid, which is 1.25 mol dm⁻³:
Volume = moles / concentration = 0.0212 mol / 1.25 mol dm⁻³ = 0.01696 dm³
Converting the volume from dm³ to cm³:
0.01696 dm³ = 0.01696 * 1000 cm³ = 16.96 cm³
Therefore, the volume of 1.25 mol dm⁻³ hydrochloric acid required to react with excess calcium carbonate to produce 227 cm³ of carbon dioxide gas measured at STP is approximately 16.96 cm³.
Based on the given answer choices, the closest option is A. 16.0 cm³.
