Question

the following unbalanced chemical equations are provided in the phet simulation: make ammonia: n₂ + h₂ → nh₃ separate water: h₂o → h₂ + o₂ combust methane: ch₄ + o₂ → co₂ + h₂o determine for which elements the specified atoms are balanced or unbalanced in each of these chemical equations when every substance in the reaction is given a coefficient of one (1). use the phet simulation to assist you in solving this problem. in the phet simulation window, click on the introduction icon at the bottom and select the bar - chart icon from the tools dropdown menu in the upper right to display how many atoms of each element are present on each side of the chemical equation. to access each chemical equation in phet, click on the radio button next to each located at the bottom of the screen. for each of the three chemical equations, set the coefficient to \1\ for each molecule. drag the appropriate items to their respective bins. atoms balanced atoms unbalanced hydrogen (separate water) hydrogen (make ammonia) carbon (combust methane) oxygen (combust methane) nitrogen (make ammonia) oxygen (separate water) hydrogen (combust methane) view available hint(s) reset help

Explanation:

Step1: Balance the combustion of methane equation

For the equation $CH_4 + O_2
ightarrow CO_2 + H_2O$, first balance the carbon atoms. There is 1 carbon atom on each side already. Then balance the hydrogen atoms. There are 4 hydrogen atoms in $CH_4$, so we need 2 moles of $H_2O$ on the product - side. Now for the oxygen atoms, we have 2 oxygen atoms in $CO_2$ and 2 oxygen atoms in 2 moles of $H_2O$, a total of 4 oxygen atoms on the product - side. So we need 2 moles of $O_2$ on the reactant - side. The balanced equation is $CH_4+2O_2
ightarrow CO_2 + 2H_2O$. Carbon: 1 on both sides, Hydrogen: 4 on both sides, Oxygen: 4 on both sides.

Step2: Balance the decomposition of water equation

For the equation $H_2O
ightarrow H_2+O_2$, we have 2 hydrogen atoms and 1 oxygen atom on the left - hand side and 2 hydrogen atoms and 2 oxygen atoms on the right - hand side. To balance the oxygen atoms, we need 2 moles of $H_2O$ on the left - hand side, which gives us 4 hydrogen atoms. So we need 2 moles of $H_2$ on the right - hand side. The balanced equation is $2H_2O
ightarrow 2H_2+O_2$. Hydrogen: 4 on both sides, Oxygen: 2 on both sides.

Step3: Balance the formation of ammonia equation

For the equation $N_2 + H_2
ightarrow NH_3$, we have 2 nitrogen atoms on the left - hand side, so we need 2 moles of $NH_3$ on the right - hand side. Now we have 6 hydrogen atoms on the right - hand side, so we need 3 moles of $H_2$ on the left - hand side. The balanced equation is $N_2+3H_2
ightarrow 2NH_3$. Nitrogen: 2 on both sides, Hydrogen: 6 on both sides.

Answer:

For Combust Methane ($CH_4 + 2O_2
ightarrow CO_2+2H_2O$): Carbon is balanced, Hydrogen is balanced, Oxygen is balanced.
For Separate Water ($2H_2O
ightarrow 2H_2 + O_2$): Hydrogen is balanced, Oxygen is balanced.
For Make Ammonia ($N_2+3H_2
ightarrow 2NH_3$): Nitrogen is balanced, Hydrogen is balanced.