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.
Is there a picture? If not then it’s probably precipitate
Hey mate here is pratyush for your help from India
..
your answer is in the attachment ..
the answer is 68g of NH3 will be produced.
hope it helps you.
be brainly