Calculate the number of moles of NO3− ions in 3.80 L of 0.200 M Al(NO3)3.?

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Calculate the number of moles of NO3− ions in 3.80 L of 0.200 M Al(NO3)3.?

Enter a numeric answer only, no units.

2006-12-01 14:50:08 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry

3 answers

the moles of Al(NO3)3 : 3.8 *0.2 = 0.76
the moles of [NO3 -] : 0.76 * 3 = 2.28

2006-12-01 15:01:31 · answer #1 · answered by James Chan 4 · 0⤊ 0⤋

213g of Al(NO3)3 make 1mole.

161.88g in 3800ml will make 0.2M Al(NO3)3.

Which is nothing but 0.76moles of Al(NO3)3

there are three (NO3)-.

Therefore, 3 x 0.76 = 2.28 moles of (NO3)-

2006-12-01 15:49:05 · answer #2 · answered by pal 1 · 0⤊ 0⤋

2.28

2006-12-01 15:10:07 · answer #3 · answered by Benny C 1 · 0⤊ 0⤋