Question

type the numbers that will balance this equation. if no number is required type a 1 in that blank. na₂co₃(aq) + bacl₂(aq) -----> baco₃(s) + nacl(aq)

Explanation:

Step1: Balance sodium atoms

Count sodium atoms on both sides. On the left, in \(Na_2CO_3\), there are 2 sodium atoms. On the right, in \(NaCl\), sodium is present. Let the coefficient of \(Na_2CO_3\) be \(a\), coefficient of \(BaCl_2\) be \(b\), coefficient of \(BaCO_3\) be \(c\), and coefficient of \(NaCl\) be \(d\). For sodium: \(2a=d\).

Step2: Balance carbonate \((CO_3^{2 -})\) and barium \((Ba^{2+})\) ions

For carbonate ions: The number of \(CO_3^{2 -}\) ions on the left (from \(Na_2CO_3\)) and on the right (from \(BaCO_3\)) must be equal. So \(a = c\). For barium ions: The number of \(Ba^{2+}\) ions on the left (from \(BaCl_2\)) and on the right (from \(BaCO_3\)) must be equal. So \(b = c\).

Step3: Balance chloride \((Cl^{-})\) ions

On the left, in \(BaCl_2\), there are \(2b\) chloride ions. On the right, in \(NaCl\), there are \(d\) chloride ions. Since \(d = 2a\) and \(a=b\) (from previous steps), if we take \(a = 1\), then \(b = 1\), \(c = 1\), and \(d=2\).

Answer:

1, 1, 1, 2