Question

i. nh4no3
ii. (naoh4)2co3
iii. al(no3)3
iv. coo(no3)2
v. (noh4)3po4
which of these formulas contain equal numbers of oxygen atoms?
a formulas ii and iii
b formulas iv and v
c formulas i and iii
d formulas i, ii, and iv

Explanation:

Step1: Count O - atoms in I

In $NH_4NO_3$, there are 3 oxygen atoms.

Step2: Count O - atoms in II

In $(NaOH)_2CO_3$, there are 2 (from $NaOH$) + 3 (from $CO_3$) = 5 oxygen atoms.

Step3: Count O - atoms in III

In $Al(NO_3)_3$, there are 3×3 = 9 oxygen atoms.

Step4: Count O - atoms in IV

In $CoO(NO_3)_2$, there are 1 (from $CoO$)+2×3 = 7 oxygen atoms.

Step5: Count O - atoms in V

In $(NOH_4)_3PO_4$, there are 3 (from $NOH_4$) + 4 (from $PO_4$)=7 oxygen atoms.

Answer:

B. Formulas IV and V