What is the number of moles in 432g Ba(NO3)2? (Ba = 137.3amu; N = 14.0amu; O = 16.0amu)
a) 0.237mol
b) 0.605mol
c) 1.65mol
d) 3.66mol
2006-12-04 20:27:20 · 5 answers · asked by me 1 in Science & Mathematics ➔ Chemistry
432 / 261 = 1.655 moles
2006-12-04 20:33:47 · answer #1 · answered by James Chan 4 · 0⤊ 0⤋
The RMM (Relative Molecular Mass) of Ba(NO3)2
= RAM (Relative Atomic Mass) of Barium + 2 x RAM of Nitrogen + 2 x 3 x RAM of Oxygen
= 137.3 + 2x14 + 6x16
= 261.3
So, the no. of moles in 432 g of Ba(NO3)2
= Grams of Ba(NO3)2 divided by RMM of Ba(NO3)2
= 432g/261.3
= 1.6532...
which is around 1.65mol.
So, the answer is c)
2006-12-05 04:35:08 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Ba(NO3)2 has Mr 261
Moles=mass / molar mass = 432 / 261 = 1.6552 so c)
2006-12-05 06:00:06 · answer #3 · answered by draco_mortifer 2 · 0⤊ 0⤋
the answer is c. 1.65 mol
2006-12-05 04:34:08 · answer #4 · answered by Dr. A 3 · 0⤊ 0⤋
answer c
2006-12-05 05:02:26 · answer #5 · answered by andrew b 2 · 0⤊ 0⤋