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reactants products h 4 h 4 o 2 o 2 __h₂ + __o₂ → __h₂o 2 reactants prod…

Question

reactants products
h 4 h 4
o 2 o 2
h₂ + o₂ → __h₂o
2 reactants products
p p
o o
p₄ + o₂ → __p₂o₃
©2015 adventures in science

Explanation:

Step1: Balance hydrogen atoms

For the reaction \(H_2 + O_2
ightarrow H_2O\), in \(H_2\) there are 2 H atoms per molecule, and in \(H_2O\) there are also 2 H atoms per molecule. Given 4 H atoms in reactants (from \(4H_2\)), in products, for \(H_2O\), since each \(H_2O\) has 2 H atoms, the coefficient of \(H_2O\) is \(\frac{4}{2}=2\). But wait, let's check oxygen. In reactants, from \(2O_2\) (2 molecules, each with 2 O atoms) there are \(2\times2 = 4\) O atoms. In products, each \(H_2O\) has 1 O atom. If the coefficient of \(H_2O\) is 4 (because \(4H_2O\) has \(4\times1=4\) O atoms). So the balanced equation is \(4H_2+2O_2 = 4H_2O\)

Step2: Balance phosphorus and oxygen atoms

For the reaction \(P_4+O_2
ightarrow P_2O_3\). In \(P_4\) there are 4 P atoms per molecule, and in \(P_2O_3\) there are 2 P atoms per molecule. Let the coefficient of \(P_4\) be \(x\) and of \(P_2O_3\) be \(y\). For P - atoms: \(4x = 2y\), assume \(x = 1\), then \(y=2\). For O - atoms: in \(O_2\) (let the coefficient be \(z\)) and \(P_2O_3\) (coefficient \(y = 2\), each \(P_2O_3\) has 3 O atoms). So \(2z=2\times3\), \(z = 3\). The balanced equation is \(P_4+3O_2 = 2P_2O_3\)

Answer:

  1. \(4H_2+2O_2 = 4H_2O\)
  2. \(P_4+3O_2 = 2P_2O_3\)