balancing equations practice day2\nbalance the following equations:\n1) __ c2h4o2 + __ o2 => __ co2 + __…

Explanation:

Step1: Balance carbon atoms in $C_2H_4O_2 + O_2\rightarrow CO_2 + H_2O$

Set coefficient of $CO_2$ to 2 to balance carbon. So we have $C_2H_4O_2 + O_2\rightarrow 2CO_2 + H_2O$.

Step2: Balance hydrogen atoms

Set coefficient of $H_2O$ to 2 to balance hydrogen. Now we have $C_2H_4O_2 + O_2\rightarrow 2CO_2+2H_2O$.

Step3: Balance oxygen atoms

Count oxygen atoms on right - hand side: $2\times2 + 2\times1=6$. On left - hand side, there are 2 in $C_2H_4O_2$, so coefficient of $O_2$ is 2. The balanced equation is $1C_2H_4O_2 + 2O_2\rightarrow 2CO_2+2H_2O$.

Step4: Balance $S_8 + O_2\rightarrow SO_2$

Set coefficient of $SO_2$ to 8 to balance sulfur. So we have $S_8 + O_2\rightarrow 8SO_2$. Then set coefficient of $O_2$ to 8 to balance oxygen. The balanced equation is $1S_8 + 8O_2\rightarrow 8SO_2$.

Step5: Balance $V_2O_5+CaS\rightarrow CaO + V_2S_5$

Set coefficient of $CaS$ to 5 to balance sulfur and calcium. We get $V_2O_5 + 5CaS\rightarrow 5CaO+V_2S_5$.

Step6: Balance $NaNO_3+PbO\rightarrow Pb(NO_3)_2 + Na_2O$

Set coefficient of $NaNO_3$ to 2 to balance nitrate and sodium. The balanced equation is $2NaNO_3+PbO\rightarrow Pb(NO_3)_2 + 1Na_2O$.

Step7: Balance $AgI+Fe_2(CO_3)_3\rightarrow FeI_3 + Ag_2CO_3$

Set coefficient of $AgI$ to 6 to balance silver and iodine, and coefficient of $Ag_2CO_3$ to 3 to balance silver and carbonate. Also, coefficient of $FeI_3$ is 2 to balance iron and iodine. The balanced equation is $6AgI + 1Fe_2(CO_3)_3\rightarrow 2FeI_3+3Ag_2CO_3$.

Step8: Balance $Mn(NO_2)_2+BeCl_2\rightarrow Be(NO_2)_2+MnCl_2$

The equation is already balanced, so the coefficients are 1, 1, 1, 1.

Answer:

1. $1C_2H_4O_2 + 2O_2\rightarrow 2CO_2+2H_2O$
2. $1S_8 + 8O_2\rightarrow 8SO_2$
3. $1V_2O_5 + 5CaS\rightarrow 5CaO+V_2S_5$
4. $2NaNO_3+PbO\rightarrow Pb(NO_3)_2 + 1Na_2O$
5. $6AgI + 1Fe_2(CO_3)_3\rightarrow 2FeI_3+3Ag_2CO_3$
6. $1Mn(NO_2)_2+1BeCl_2\rightarrow 1Be(NO_2)_2+1MnCl_2$