Balancing Chemical Equations Balance the equation Na(s) + H20(/)—? NaOH(aq) + H2 (g) The unbalanced equation for the reaction between methane and bromine is Once this equation is balanced what is the value of the coefficient in front of bromine Br2? (a)1, (b) 2, (c) 3, (d) 4, (e) 6.

Answer Begin by counting each kind of atom on the two sides of the arrow. There are one Na, one O, and two H on the left side, and one Na, one O, and three H on the right. The Na and O atoms are balanced, but the number of H atoms is not. To increase the number of H atoms on the left, let’s try placing the coefficient 2 in front of H2O: Na(s)+2H2O(l) NaOH(aq)+H2(g). Although beginning this way does not balance H, it does increase the number of reactant H atoms, which we need to do. (Also, adding the coefficient 2 on H2O unbalanced O, but we will take care of that after we balance H.) Now that we have 2 H2O on the left, we balance H by putting the coefficient 2 in front of NaOH: Na (aq)+2H2O2NaOH+H2 Balancing H in this way brings O into balance, but now Na is unbalanced, with one Na on the left and two on the right. To rebalance Na, we put the coefficient 2 in front of the reactant: 2Na+2H2O 2NaOH + H2 We now have two Na atoms, four H atoms, and two O atoms on each side. The equation is balanced. Comment Notice that we moved back and forth, placing a coefficient in front of H2O, then NaOH, and finally Na. In balancing equations, we often find ourselves following this pattern of moving back and forth from one side of the arrow to the other, placing coefficients first in front of a formula on one side...