balancing equations redo\n1. bao₂ → bao + o₂\n2. ch₄ + cl₂ → chcl₃ + hcl\n3. __sbcl₃ + __h₂s → __sb₂s₃ +…

balancing equations redo\n1. bao₂ → bao + o₂\n2. ch₄ + cl₂ → chcl₃ + hcl\n3. __sbcl₃ + __h₂s → __sb₂s₃ + __hcl\n4. mgcl₂ + naoh → mg(oh)₂ + nacl\n5. __h₂o + __co₂ → __c₇h₈ + __o₂

balancing equations redo\n1. bao₂ → bao + o₂\n2. ch₄ + cl₂ → chcl₃ + hcl\n3. __sbcl₃ + __h₂s → __sb₂s₃ + __hcl\n4. mgcl₂ + naoh → mg(oh)₂ + nacl\n5. __h₂o + __co₂ → __c₇h₈ + __o₂

Answer

Explanation:

Step1: Balance oxygen in the first equation

We have 2 oxygen atoms in $BaO_2$ on the left - hand side and 3 oxygen atoms on the right - hand side ($1$ in $BaO$ and $2$ in $O_2$). Multiply $BaO_2$ and $BaO$ by 2 to get $2BaO_2\rightarrow2BaO + O_2$.

Step2: Balance chlorine in the second equation

On the left - hand side, we have 2 chlorine atoms in $Cl_2$ and on the right - hand side, 4 chlorine atoms in $CHCl_3$ and $HCl$. Multiply $Cl_2$ by 4, $CH_4$ by 1, $CHCl_3$ by 1 and $HCl$ by 3 to get $CH_4+4Cl_2\rightarrow CHCl_3 + 3HCl$.

Step3: Balance antimony and sulfur in the third equation

For antimony, we need 2 moles of $SbCl_3$ to get 2 antimony atoms on the left - hand side to match $Sb_2S_3$ on the right - hand side. For sulfur, we need 3 moles of $H_2S$ to match 3 sulfur atoms in $Sb_2S_3$. Then we have 6 moles of $HCl$ on the right - hand side. So the equation is $2SbCl_3+3H_2S\rightarrow Sb_2S_3 + 6HCl$.

Step4: Balance magnesium, chlorine, sodium and oxygen in the fourth equation

We have 1 magnesium atom on both sides. For chlorine, we have 2 chlorine atoms in $MgCl_2$, so we need 2 moles of $NaCl$. For sodium and oxygen, we need 2 moles of $NaOH$. The balanced equation is $MgCl_2+2NaOH\rightarrow Mg(OH)_2 + 2NaCl$.

Step5: Balance carbon, hydrogen and oxygen in the fifth equation

For carbon, we have 1 carbon atom in $CO_2$ and 7 carbon atoms in $C_7H_8$. So we need 7 moles of $CO_2$. For hydrogen, we have 2 hydrogen atoms in $H_2O$ and 8 hydrogen atoms in $C_7H_8$. So we need 4 moles of $H_2O$. For oxygen, we have $4 + 14=18$ oxygen atoms on the left - hand side and we need 9 moles of $O_2$ on the right - hand side. The balanced equation is $4H_2O+7CO_2\rightarrow C_7H_8 + 9O_2$.

Answer:

  1. $2BaO_2\rightarrow2BaO + O_2$
  2. $CH_4+4Cl_2\rightarrow CHCl_3 + 3HCl$
  3. $2SbCl_3+3H_2S\rightarrow Sb_2S_3 + 6HCl$
  4. $MgCl_2+2NaOH\rightarrow Mg(OH)_2 + 2NaCl$
  5. $4H_2O+7CO_2\rightarrow C_7H_8 + 9O_2$