Answer is: <span>A. 18.02 g/mol.
At standard temperature and pressure 1 mol of gas occupied 22.4 liters:
V(H</span>₂O) = 22.4 L; volume of water.
Vm = 22.4 L/mol; molar volume at STP.
n(H₂O) = V ÷ Vm.
n(H₂O) = 22.4 L ÷ 22.4 L/mol.
n(H₂O) = 1 mol; amount of substance (water).
M(H₂O) = Ar(O) + 2Ar(H) · g/mol.
M(H₂O) = 16 + 2 ·1.01 · g/mol.
M(H₂O) = 18.02 g/mol; molar mass of water.
If excess carbon disulfide reacts with 450 mL of oxygen, 150 mL of carbon dioxide and 300 mL of sulfur dioxide gases would be produced respectively.
<h3>Stoichiometric calculation</h3>
The reaction between liquid carbon disulfide and oxygen is represented by the equations below:

The mole ratio of oxygen to carbon dioxide and sulfur dioxide produced is 3:1:2.
Thus, for 450 mL oxygen, 1/3 x 450 = 150 mL of carbon dioxide will be required.
Also for 450 mL of oxygen, 2/3 x 450 = 300 mL of sulfur dioxide will be required.
More on stoichiometric calculations can be found here: brainly.com/question/27287858
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I would say chemical but I’m not 100% sure might wanna get a second opinion