Answer:
Step-by-step explanation:
To determine the amount of chlorine needed to make 75.0 grams of C2H2Cl4, we need to balance the chemical equation and then use the molar ratio between the reactants and product to find the amount of Cl2 needed.
The balanced equation is:
2 Cl2(g) + C2H2(g) -> C2H2Cl4(l)
Next, we can use the molar mass of C2H2Cl4 to find the number of moles of C2H2Cl4 produced:
75.0 g C2H2Cl4 x (1 mole C2H2Cl4 / 153.8 g C2H2Cl4) = 0.489 moles C2H2Cl4
Since the balanced equation has a 1:2 ratio of C2H2 to Cl2, this means that we need 2 moles of Cl2 for every mole of C2H2Cl4 produced. Therefore, we will need 2 x 0.489 moles = 0.978 moles of Cl2.
Finally, to find the volume of Cl2 at STP, we can use the ideal gas law:
PV = nRT
Where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant (0.0821 L-atm/mol-K), and T is the temperature in Kelvin (273 K).
Since the pressure is 1 atm and the temperature is 273 K, we can rearrange the equation to solve for volume:
V = nRT / P = 0.978 moles x 0.0821 L-atm/mol-K x 273 K / 1 atm = 17.1 L
Therefore, 17.1 liters of Cl2 at STP will be needed to make 75.0 grams of C2H2Cl4.