To determine the mass of [tex]\( H_2O \)[/tex] formed during the reaction described by the equation:
[tex]\[ CH_4 + O_2 \longrightarrow CO_2 + H_2O \][/tex]
we use the Law of Conservation of Mass, which states that the total mass of reactants should equal the total mass of products.
Step-by-Step Solution:
1. Identify the masses of reactants:
- Mass of [tex]\( CH_4 \)[/tex]: 16 grams
- Mass of [tex]\( O_2 \)[/tex]: 64 grams
Therefore, the total mass of reactants is:
[tex]\[
\text{Total mass of reactants} = 16 \, \text{g (from } CH_4) + 64 \, \text{g (from } O_2) = 80 \, \text{g}
\][/tex]
2. Identify the mass of one of the products:
- Mass of [tex]\( CO_2 \)[/tex]: 44 grams
3. Apply the Law of Conservation of Mass:
Since the total mass of products must equal the total mass of reactants, we have:
[tex]\[
\text{Total mass of products} = \text{Total mass of reactants} = 80 \, \text{g}
\][/tex]
4. Determine the mass of [tex]\( H_2O \)[/tex]:
Let [tex]\( m \)[/tex] be the mass of [tex]\( H_2O \)[/tex]. According to the reaction, the total mass of products is the sum of the masses of [tex]\( CO_2 \)[/tex] and [tex]\( H_2O \)[/tex]:
[tex]\[
44 \, \text{g (from } CO_2) + m \, \text{g (from } H_2O) = 80 \, \text{g}
\][/tex]
Solving for [tex]\( m \)[/tex]:
[tex]\[
m = 80 \, \text{g} - 44 \, \text{g} = 36 \, \text{g}
\][/tex]
Therefore, the mass of [tex]\( H_2O \)[/tex] formed during the reaction is [tex]\( \boxed{36 \, \text{grams}} \)[/tex].
