steps for balancing equations\n__hgo→__hg + __o₂\nhg - 1\no - 1\nhg - 1\no - 2\n2al+3cucl₂→3cu + 2alcl₃\nal…
steps for balancing equations\n__hgo→__hg + __o₂\nhg - 1\no - 1\nhg - 1\no - 2\n2al+3cucl₂→3cu + 2alcl₃\nal - 1\nal - 1\ncu - 1\ncu - 1\ncl - 2\ncl - 3\npartner practice: __n₂+__h₂→__nh₃\nn - 2\nn - 1\nh - 2\nh - 3\npartner practice: __kclo₃→__kcl+__o₂\nk - 1\ncl - 1\no - 3\nk - 1\ncl - 1\no - 2\npartner practice: __kbr+__cl₂→__kcl+__br₂\nk - 1\nbr - 1\ncl - 2\nk - 1\ncl - 1\nbr - 2\npartner practice: __co+__o₂→__co₂\nc - 1\nc - 1\no - 2\no - 2\nsolo practice: 4c+s₈→4cs₂\nc - 1\nc - 1\ns - 8\ns - 2\nsolo practice: __p₄+__br₂→__pbr₃\np - 4\np - 1\nbr - 2\nbr - 3\nsolo practice: __na₃p+__caf₂→__naf+__ca₃p₂\nna - 1\nna - 1\np - 1\np - 1\nca - 1\nca - 3\nf - 2\nf - 1
Answer
Explanation:
Step1: Balance the mercury and oxygen in $HgO \rightarrow Hg+O_2$
We have 1 oxygen atom on the left - hand side and 2 on the right - hand side. Multiply $HgO$ by 2 to get $2HgO \rightarrow 2Hg + O_2$.
Step2: Balance $N_2 + H_2\rightarrow NH_3$
There are 2 nitrogen atoms in $N_2$ and 1 in $NH_3$, and 2 hydrogen atoms in $H_2$ and 3 in $NH_3$. Multiply $N_2$ by 1, $H_2$ by 3 and $NH_3$ by 2 to get $N_2+3H_2\rightarrow 2NH_3$.
Step3: Balance $KClO_3\rightarrow KCl + O_2$
There are 3 oxygen atoms in $KClO_3$ and 2 in $O_2$. Multiply $KClO_3$ by 2, $KCl$ by 2 and $O_2$ by 3 to get $2KClO_3\rightarrow 2KCl + 3O_2$.
Step4: Balance $KBr+Cl_2\rightarrow KCl + Br_2$
There are 2 chlorine atoms in $Cl_2$ and 1 in $KCl$, and 2 bromine atoms in $Br_2$ and 1 in $KBr$. Multiply $KBr$ by 2, $KCl$ by 2 to get $2KBr+Cl_2\rightarrow 2KCl + Br_2$.
Step5: Balance $CO + O_2\rightarrow CO_2$
There are 2 oxygen atoms in $O_2$ and 1 in $CO$. Multiply $CO$ by 2 and $CO_2$ by 2 to get $2CO+O_2\rightarrow 2CO_2$.
Step6: Balance $C + S_8\rightarrow CS_2$
There are 8 sulfur atoms in $S_8$ and 2 in $CS_2$. Multiply $CS_2$ by 4 and $C$ by 4 to get $4C + S_8\rightarrow 4CS_2$.
Step7: Balance $P_4+Br_2\rightarrow PBr_3$
There are 4 phosphorus atoms in $P_4$ and 1 in $PBr_3$. Multiply $PBr_3$ by 4. There are 2 bromine atoms in $Br_2$ and 3 in $PBr_3$. Multiply $Br_2$ by 6 to get $P_4 + 6Br_2\rightarrow 4PBr_3$.
Step8: Balance $Na_3P+CaF_2\rightarrow NaF+Ca_3P_2$
There are 3 sodium atoms in $Na_3P$ and 1 in $NaF$, 1 phosphorus atom in $Na_3P$ and 2 in $Ca_3P_2$, 1 calcium atom in $CaF_2$ and 3 in $Ca_3P_2$, 2 fluorine atoms in $CaF_2$ and 1 in $NaF$. Multiply $Na_3P$ by 2, $CaF_2$ by 3, $NaF$ by 6 to get $2Na_3P+3CaF_2\rightarrow 6NaF + Ca_3P_2$.
Answer:
$2HgO \rightarrow 2Hg + O_2$; $N_2+3H_2\rightarrow 2NH_3$; $2KClO_3\rightarrow 2KCl + 3O_2$; $2KBr+Cl_2\rightarrow 2KCl + Br_2$; $2CO+O_2\rightarrow 2CO_2$; $4C + S_8\rightarrow 4CS_2$; $P_4 + 6Br_2\rightarrow 4PBr_3$; $2Na_3P+3CaF_2\rightarrow 6NaF + Ca_3P_2$