Question

name: ahron jvare2 period/block: a5
make ammonia
chemical equation:
$n_{2}+h_{2}\to nh_{3}$
model:
$n_{2}+h_{2}\to nh_{3}$
atom counts:
final chemical equation:
$\\_\\_n_{2}+\\_\\_h_{2}\to\\_\\_nh_{3}$
separation of water
chemical equation:
$h_{2}o\to h_{2}+o_{2}$
model:
$h_{2}o\to h_{2}+o_{2}$
atom counts:
final chemical equation:
$\\_\\_h_{2}o\to\\_\\_h_{2}+\\_\\_o_{2}$

Explanation:

Step1: Balance the ammonia equation

For \(N_2 + H_2
ightarrow NH_3\), there are 2 \(N\) atoms on the left and 1 on the right. So, put a 2 in front of \(NH_3\): \(N_2 + H_2
ightarrow 2NH_3\). Now, there are 6 \(H\) atoms on the right. Put a 3 in front of \(H_2\). The balanced equation is \(N_2+3H_2
ightarrow 2NH_3\).

• Reactant \(N\): 2 (from \(N_2\)), Product \(N\): 2 (from \(2NH_3\))
• Reactant \(H\): 6 (from \(3H_2\)), Product \(H\): 6 (from \(2NH_3\))

Step2: Balance the water - separation equation

For \(H_2O
ightarrow H_2 + O_2\), there is 1 \(O\) atom on the left and 2 on the right. Put a 2 in front of \(H_2O\): \(2H_2O
ightarrow H_2+O_2\). Now, there are 4 \(H\) atoms on the left. Put a 2 in front of \(H_2\). The balanced equation is \(2H_2O
ightarrow 2H_2 + O_2\).

• Reactant \(H\): 4 (from \(2H_2O\)), Product \(H\): 4 (from \(2H_2\))
• Reactant \(O\): 2 (from \(2H_2O\)), Product \(O\): 2 (from \(O_2\))

Answer:

• For \(N_2 + H_2

ightarrow NH_3\):

• Reactant \(N\): 2, Product \(N\): 2; Reactant \(H\): 2, Product \(H\): 3. Final equation: \(1N_2 + 3H_2

ightarrow 2NH_3\)

• For \(H_2O

ightarrow H_2 + O_2\):

• Reactant \(H\): 2, Product \(H\): 2; Reactant \(O\): 1, Product \(O\): 2. Final equation: \(2H_2O

ightarrow 2H_2+1O_2\)