Question

balancing equations - questions
instructions: balance the chemical equations below. identify how many atoms of each element are present once the equation is balanced.
#1
hgo → hg + __ o2
hg: __
o: __
hg: __
o: __
#2
p4 + o2 → __ p2o3
p: __
o: __
p: __
o: __
#3
bas + li3n → ba3n2 + li2s
ba: __
s: __
li: __
n: __
ba: __
s: __
li: __
n: __
#4
naf + br2 → nabr + f2
na: __
f: __
br: __
na: __
f: __
br: __

Explanation:

Step1: Balance HgO → Hg + O₂

Oxygen is un - balanced. There are 2 oxygen atoms in O₂ on the right - hand side. So, put 2 in front of HgO. Then, there are 2 Hg atoms on the left - hand side, so put 2 in front of Hg on the right - hand side.
2HgO → 2Hg + O₂
Hg: 2 on left, 2 on right; O: 2 on left, 2 on right

Step2: Balance P₄ + O₂ → P₂O₃

First, balance the phosphorus atoms. Since there are 4 P atoms in P₄, put 2 in front of P₂O₃. Then, there are 6 O atoms in 2P₂O₃, so put 3 in front of O₂.
P₄ + 3O₂ → 2P₂O₃
P: 4 on left, 4 on right; O: 6 on left, 6 on right

Step3: Balance BaS + Li₃N → Ba₃N₂ + Li₂S

For barium, since there are 3 Ba atoms in Ba₃N₂, put 3 in front of BaS. For nitrogen, since there are 2 N atoms in Ba₃N₂, put 2 in front of Li₃N. Then, for lithium, 2Li₃N has 6 Li atoms, so put 3 in front of Li₂S. And for sulfur, 3BaS has 3 S atoms and 3Li₂S has 3 S atoms.
3BaS + 2Li₃N → Ba₃N₂ + 3Li₂S
Ba: 3 on left, 3 on right; S: 3 on left, 3 on right; Li: 6 on left, 6 on right; N: 2 on left, 2 on right

Step4: Balance NaF + Br₂ → NaBr + F₂

For fluorine, since there are 2 F atoms in F₂, put 2 in front of NaF. Then, for sodium, 2NaF has 2 Na atoms, so put 2 in front of NaBr. And for bromine, 2NaBr has 2 Br atoms and Br₂ has 2 Br atoms.
2NaF + Br₂ → 2NaBr + F₂
Na: 2 on left, 2 on right; F: 2 on left, 2 on right; Br: 2 on left, 2 on right

Answer:

#1: 2HgO → 2Hg + O₂; Hg: 2, 2; O: 2, 2
#2: P₄ + 3O₂ → 2P₂O₃; P: 4, 4; O: 6, 6
#3: 3BaS + 2Li₃N → Ba₃N₂ + 3Li₂S; Ba: 3, 3; S: 3, 3; Li: 6, 6; N: 2, 2
#4: 2NaF + Br₂ → 2NaBr + F₂; Na: 2, 2; F: 2, 2; Br: 2, 2