Final answer:
The volume of CO2 gas produced when 5g of HCl reacts with excess calcium carbonate is approximately 1.534 liters at STP, calculated using stoichiometry and the ideal gas law.
Step-by-step explanation:
To calculate the volume of gas produced when 5g HCl reacts with excess calcium carbonate, we first need to use the stoichiometry of the balanced chemical equation:
CaCO3 + 2 HCl → CaCl2 + CO2 + H2O
From the equation we can see that 2 moles of HCl produce 1 mole of CO2 gas. Now we need to find the number of moles of HCl that 5g represents, and then use this to calculate the moles of CO2 produced.
First, we calculate the molar mass of HCl (1.007 + 35.453 = 36.460 g/mol). Then, we find the number of moles in 5g:
Moles of HCl = mass (g) / molar mass (g/mol) = 5g / 36.460 g/mol = 0.137 moles HCl
Since the reaction requires 2 moles of HCl to produce 1 mole of CO2, we then calculate the moles of CO2:
Moles of CO2 = 0.137 moles HCl / 2 = 0.0685 moles CO2
Finally, to find the volume of CO2 produced at STP (standard temperature and pressure), we use the ideal gas law where 1 mole of gas at STP occupies 22.4 L:
Volume of CO2 = 0.0685 moles × 22.4 L/mole = 1.534 L
Therefore, the volume of gas produced when 5g of HCl reacts with excess calcium carbonate is approximately 1.534 liters of CO2 at STP.