Answer:
A. 30cm³
Step-by-step explanation:
Based on the chemical reaction:
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
1 mol of calcium carbonate reacts with 2 moles of HCl to produce 1 mol of CO₂
To solve this question we must convert the mass of each reactant to moles. With the moles we can find limiting reactant and the moles of CO₂ produced. Using PV = nRT we can find the volume of the gas:
Moles CaCO₃ -Molar mass: 100.09g/mol-
1.00g * (1mol / 100.09g) = 9.991x10⁻³ moles
Moles HCl:
50cm³ = 0.0500dm³ * (0.05 mol / dm³) = 2.5x10⁻³ moles
For a complete reaction of 2.5x10⁻³ moles HCl there are necessaries:
2.5x10⁻³ moles HCl * (1mol CaCO₃ / 2mol HCl) = 1.25x10⁻³ moles CaCO₃. As there are 9.991x10⁻³ moles, HCl is limiting reactant.
The moles produced of CO₂ are:
2.5x10⁻³ moles HCl * (1mol CO₂ / 2mol HCl) = 1.25x10⁻³ moles CO₂
Using PV = nRT
Where P is pressure = 1atm assuming STP
V volume in L
n moles = 1.25x10⁻³ moles CO₂
R gas constant = 0.082atmL/molK
T = 273.15K at STP
V = nRT / P
1.25x10⁻³ moles * 0.082atmL/molK*273.15K / 1atm = V
0.028L = V
28cm³ = V
As 28cm³ ≈ 30cm³
Right option is:
A. 30cm³