AlCl3(aq) + 4NaOH(aq) → NaAlO2 (aq) + 3NaCl(aq) +2H2O(l)?
Also how did you derive at this answer?
Are there conversion factors involved?
2007-01-30 08:29:57 · 3 answers · asked by San Fran Kid 2 in Science & Mathematics ➔ Chemistry
▲The factors could be ignored coz 1 atom of Al per molecule on LHS goes into 1atom of Al per molecule on RHS, being other stuff Al free.
▲here comes proportion: 4.46/m1 = x/m2, hence x=4.46g*(m2/m1), where m1 is atomic weight of AlCl3, m2 is atomic weight of NaAlO2. Sorry, got no Mendeleev table at hand.
2007-01-30 09:04:41 · answer #1 · answered by Anonymous · 0⤊ 0⤋
just 1 eq of your AlCl3.
Also how did you derive at this answer?
man, just read the equation !
Are there conversion factors involved?
i told you , the factor is 1 ---> 1 eq
now do the calculation yourself
2007-01-30 16:36:58 · answer #2 · answered by scientific_boy3434 5 · 0⤊ 0⤋
This looks like a homework problem or maybe a question on a test.
2007-01-30 16:36:03 · answer #3 · answered by jim m 5 · 1⤊ 0⤋