Question

problem #5: _mgf₂ + _li₂co₃ → _mgco₃ + _lif
step 1: count
mg=
f=
li=
c=
o=
mg=
f=
li=
c=
o=
step 2: circle- balanced or unbalanced
step 3: if unbalanced then balance it

Explanation:

Step1: Count atoms

Left side: \(Mg = 1\), \(F = 2\), \(Li = 2\), \(C = 1\), \(O = 3\)
Right side: \(Mg = 1\), \(F = 1\), \(Li = 1\), \(C = 1\), \(O = 3\)

Step2: Balance \(F\) and \(Li\)

Multiply \(LiF\) by \(2\) to balance \(F\). Now \(Li\) on right side is \(2\). Since \(Li_2CO_3\) has \(2\) \(Li\) atoms, coefficients for \(MgF_2\), \(Li_2CO_3\), \(MgCO_3\), \(LiF\) are \(1\), \(1\), \(1\), \(2\)

Answer:

\(1MgF_2+1Li_2CO_3
ightarrow1MgCO_3 + 2LiF\)