4) choose the best answer.
what is the coefficient of o₂ in experiment 1?
2
3
1
gas
5) choose the best answer.
hydrogen cyanide (hcn) and water (h₂o) are produced from the reaction of methane (ch₄), ammonia (nh₃), and oxygen gas (o₂).
which of these best represents the reactants for this reaction?
hcn + h₂o + ch₄
ch₄ + nh₃ + o₂
hcn + h₂o
ch₄ + nh₃ + o

Question

Accepted Answer

### Question 4
# Brief Explanations:
To find the coefficient of $ O_2 $ in Experiment 1 (assuming the reaction is related to the production of HCN and $ H_2O $ from $ CH_4 $, $ NH_3 $, and $ O_2 $), we balance the reaction: $ 2CH_4 + 2NH_3 + 3O_2
ightarrow 2HCN + 6H_2O $. From this balanced equation, the coefficient of $ O_2 $ is 3. But since the options include 2, 3, 1, gas, and assuming the context (maybe a specific experiment), if we consider a possible simpler or mis - balanced (but common in basic problems) or the given options, if the reaction was $ CH_4+NH_3 + O_2
ightarrow HCN + H_2O $, balancing C: 1, N:1, H: $ 4 + 3=7 $ in reactants, $ 1+2 = 3 $ in products. Wait, no, the correct balanced reaction for $ CH_4+NH_3 + O_2
ightarrow HCN + H_2O $ is:

1. Balance C: $ CH_4
ightarrow HCN $, so 1 C on each side.
2. Balance N: $ NH_3
ightarrow HCN $, so 1 N on each side.
3. Balance H: Reactants: $ 4 (from CH_4)+3 (from NH_3)=7 $, Products: $ 1 (from HCN)+2x (from H_2O) $. Let $ x $ be the coefficient of $ H_2O $. So $ 1 + 2x=7\Rightarrow x = 3 $.
4. Balance O: Products: $ 3 (from 3H_2O)$ O atoms. Reactants: $ O_2 $, so coefficient of $ O_2=\frac{3}{2} $, multiply all by 2: $ 2CH_4+2NH_3 + 3O_2
ightarrow 2HCN + 6H_2O $. But if the experiment has a different setup, but among the options, 3 is a possible coefficient for $ O_2 $ in the balanced reaction. However, if we assume a different approach, maybe the question has a typo or a simpler reaction. But from the options, the most probable answer (if we consider the balanced reaction for the production of HCN and $ H_2O $ from $ CH_4 $, $ NH_3 $, and $ O_2 $) is 3. But wait, maybe the experiment 1 has a reaction where the coefficient of $ O_2 $ is 3. But if we made a mistake, and the reaction was $ CH_4+NH_3+O_2
ightarrow HCN + H_2O $ with a wrong balance, but the options have 3 as an option. Alternatively, if the reaction was $ 2CH_4 + 2NH_3+3O_2 = 2HCN + 6H_2O $, then $ O_2 $ has coefficient 3.
# Answer:
3 (assuming the context of the reaction between $ CH_4 $, $ NH_3 $, and $ O_2 $ to form $ HCN $ and $ H_2O $ and the given options)

### Question 5
# Brief Explanations:
Reactants are the substances that react together to form products. The problem states that hydrogen cyanide (HCN) and water ($ H_2O $) are produced from the reaction of methane ($ CH_4 $), ammonia ($ NH_3 $), and oxygen gas ($ O_2 $). So the reactants are the starting materials, which are $ CH_4 $, $ NH_3 $, and $ O_2 $.

- Option 1: $ HCN + H_2O+CH_4 $: $ HCN $ and $ H_2O $ are products, not reactants.
- Option 2: $ CH_4 + NH_3+O_2 $: These are the reactants as per the problem statement.
- Option 3: $ HCN + H_2O $: These are products, not reactants.
- Option 4: $ CH_4 + NH_3+O $: Oxygen is diatomic ($ O_2 $) in the reaction, not atomic O.

# Answer:
B. $ CH_4 + NH_3+O_2 $ (using the option identifier as per the original, if the options were labeled A: $ HCN + H_2O+CH_4 $, B: $ CH_4 + NH_3+O_2 $, C: $ HCN + H_2O $, D: $ CH_4 + NH_3+O $)

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