QUESTION IMAGE
Question
independent practice -- 8 problems (spa coefficients)
- h₂ + cl₂ → __ hcl
- c + o₂ → co₂
- al + o₂ → __ al₂o₃
- kclo₃ → kcl + __ o₂
- ca(oh)₂ + hcl → cacl₂ + h₂o
- fe + __ hcl → fecl₂ + h₂
- c₃h₆ + o₂ → co₂ + h₂o
- p₄ + o₂ → p₂o₅
Step1: Balance the first equation
For \(H_2 + Cl_2 \to HCl\), count the number of \(H\) and \(Cl\) atoms. There are 2 \(H\) and 2 \(Cl\) on the left. So, we need 2 \(HCl\) on the right.
\(H_2+Cl_2 = 2HCl\)
Step2: Check the second equation
For \(C + O_2\to CO_2\), there is 1 \(C\) and 2 \(O\) on the left, and 1 \(C\) and 2 \(O\) on the right. It is already balanced.
\(C + O_2=CO_2\)
Step3: Balance the third equation
For \(Al+O_2\to Al_2O_3\), use the least - common multiple method. The LCM of 2 (from \(O_2\)) and 3 (from \(Al_2O_3\)) for \(O\) atoms is 6. So, we need \(3O_2\) and \(2Al_2O_3\). Then balance \(Al\) atoms: \(4Al + 3O_2=2Al_2O_3\)
Step4: Balance the fourth equation
For \(KClO_3\to KCl + O_2\), the LCM of 2 (from \(O_2\)) and 3 (from \(KClO_3\)) for \(O\) atoms is 6. So, we need \(2KClO_3\) and \(3O_2\). Then balance \(K\) and \(Cl\) atoms: \(2KClO_3 = 2KCl+3O_2\)
Step5: Balance the fifth equation
For \(Ca(OH)_2+HCl\to CaCl_2 + H_2O\), balance \(Cl\) atoms first. We need \(2HCl\). Then balance \(H\) and \(O\) atoms: \(Ca(OH)_2 + 2HCl=CaCl_2+2H_2O\)
Step6: Balance the sixth equation
For \(Fe + HCl\to FeCl_2+H_2\), balance \(Cl\) atoms. We need \(2HCl\). Then balance \(H\) atoms: \(Fe + 2HCl=FeCl_2+H_2\)
Step7: Balance the seventh equation
For \(C_3H_6+O_2\to CO_2 + H_2O\), balance \(C\) atoms first: \(C_3H_6+O_2\to 3CO_2+H_2O\). Then balance \(H\) atoms: \(C_3H_6+O_2\to 3CO_2 + 3H_2O\). Now balance \(O\) atoms: \(2C_3H_6+9O_2 = 6CO_2+6H_2O\)
Step8: Balance the eighth equation
For \(P_4+O_2\to P_2O_5\), balance \(P\) atoms first: \(P_4+O_2\to 2P_2O_5\). Then balance \(O\) atoms: \(P_4 + 5O_2=2P_2O_5\)
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