If 8.29 L of gaseous Cl2 measured at STP and 0.380 mol of liquid CS2 are reacted stoichiometrically according to the balanced equation, how many moles of liquid CS2 remain?

CS2(l) + 3Cl2(g) ¡æ CCl4(l) + S2Cl2(l)

Molar Mass (g/mol)
Cl2 70.906
CS2 76.143

CS2 1.26
Molar Volume (L)
22.4 at STP

Gas Constant
(L.atm.mol-1.K-1)
0.0821

2+2=4-1=3 quick maffs

To determine how many moles of liquid CS2 remain after the reaction, we need to use stoichiometry. Stoichiometry is the calculation of reactants and products in a chemical reaction based on the balanced equation.

Let's start by looking at the balanced equation:
CS2(l) + 3Cl2(g) → CCl4(l) + S2Cl2(l)

From the balanced equation, we can see that 1 mole of CS2 reacts with 3 moles of Cl2.

Given:
Volume of Cl2 gas = 8.29 L (measured at STP)
Molar volume at STP = 22.4 L/mol (which means 22.4 L of any gas at STP contains 1 mole of that gas)

To determine the number of moles of Cl2 gas, we can use the molar volume at STP:

moles of Cl2 = Volume of Cl2 gas / Molar volume at STP

moles of Cl2 = 8.29 L / 22.4 L/mol
moles of Cl2 = 0.370 mol

Given:
Moles of CS2 = 0.380 mol

Since the stoichiometric ratio between Cl2 and CS2 is 3:1, we need to compare the moles of Cl2 to the moles of CS2.

The ratio of moles of Cl2 to CS2 is 3:1, which means for every 3 moles of Cl2, we need 1 mole of CS2.

So, if the ratio is 3:1, with 0.370 moles of Cl2, the corresponding moles of CS2 used in the reaction would be:

moles of CS2 used = (moles of Cl2) × (1 mole of CS2 / 3 moles of Cl2)
moles of CS2 used = 0.370 mol × (1 mol / 3 mol)
moles of CS2 used = 0.123 mol

To find the moles of CS2 remaining, we subtract the moles of CS2 used from the initial moles of CS2:

moles of CS2 remaining = initial moles of CS2 - moles of CS2 used
moles of CS2 remaining = 0.380 mol - 0.123 mol
moles of CS2 remaining = 0.257 mol

Therefore, 0.257 moles of liquid CS2 remain after the reaction.