Step-by-step explanation:
To determine how many liters of ethyne (C₂H₂) can form at STP, we need to use stoichiometry and the ideal gas law.
First, let's determine the number of moles of CaC₂ and H₂O that reacted. From the given information, we know that the mass of water is 10.0 g and the mass of CaC₂ is 2.0 g. We can use the molar masses of CaC₂ (64.10 g/mol) and H₂O (18.02 g/mol) to convert the masses to moles:
moles of CaC₂ = 2.0 g / 64.10 g/mol = 0.0312 mol
moles of H₂O = 10.0 g / 18.02 g/mol = 0.555 mol
Next, we need to determine the limiting reactant to find out how many moles of ethyne can form. To do this, we compare the mole ratio of CaC₂ to H₂O in the balanced equation:
CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂
1 2
For every 1 mole of CaC₂, we need 2 moles of H₂O to react. The moles of H₂O we have (0.555 mol) are more than enough to react with the moles of CaC₂ we have (0.0312 mol). Therefore, CaC₂ is the limiting reactant.
Now we can use the mole ratio of CaC₂ to C₂H₂ in the balanced equation to find the moles of C₂H₂ that can form:
CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂
1 1
1 mole of CaC₂ produces 1 mole of C₂H₂. Therefore, the moles of C₂H₂ that can form is also 0.0312 mol.
Finally, we can use the ideal gas law to find the volume of C₂H₂ at STP. The ideal gas law is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. At STP, the pressure is 1 atm and the temperature is 273 K.
Plugging in the values, we get:
V = (nRT) / P
V = (0.0312 mol)(0.0821 L·atm/mol·K)(273 K) / 1 atm
V = 0.738 L
Therefore, the volume of C₂H₂ that can form at STP is 0.738 L.