To answer the question above, multiply the given number of moles by the molar masses.
(A) (0.20 mole) x (32 g / 1 mole) = 6.4 grams O2
(B) (0.75 mole) x (62 g / 1 mole) = 46.5 grams H2CO3
(C) (3.42 moles) x (28 g / 1 mole) = 95.7 grams CO
(D) (4.1 moles) x (29.88 g / 1 mole) = 122.508 g Li2O
The answer to the question above is letter D.
