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.
Answer:
120g
Explanation:
Step 1:
We'll begin by writing the balanced equation for the reaction.
Sn + 2HF —> SnF2 + H2
Step 2:
Determination of the number of mole HF needed to react with 3 moles of Sn.
From the balanced equation above,
1 mole of Sn and reacted with 2 moles of HF.
Therefore, 3 moles Sn will react with = 3 x 2 = 6 moles of HF.
Step 3:
Conversion of 6 moles of HF to grams.
Number of mole HF = 6 moles
Molar Mass of HF = 1 + 19 = 20g/mol
Mass of HF =..?
Mass = number of mole x molar Mass
Mass of HF = 6 x 20
Mass of HF = 120g
Therefore, 120g of HF is needed to react with 3 moles of Sn.
Answer chocies .......................
