Answer :
Certainly! To determine which equation has water (H₂O) as a reactant in the chemical reaction, we'll analyze each equation:
A. [tex]\( Na + H_2O \rightarrow NaOH + H_2 \)[/tex]
In this equation, [tex]\( H_2O \)[/tex] is clearly present on the left-hand side of the reaction, which means it is a reactant.
B. [tex]\( H_2 + O_2 \rightarrow H_2O \)[/tex]
In this equation, [tex]\( H_2O \)[/tex] is found on the right-hand side, which means it is a product, not a reactant.
C. [tex]\( CO_2 + O_2 + \text{energy} \rightarrow C_6H_{12}O_6 + O_2 \)[/tex]
Here, [tex]\( H_2O \)[/tex] does not appear on either side of the equation, meaning it is neither a reactant nor a product.
D. [tex]\( Na + Cl \rightarrow NaCl \)[/tex]
In this equation, [tex]\( H_2O \)[/tex] is also not present on either side of the reaction.
Based on this analysis, the equation that has water (H₂O) as a reactant is:
A. [tex]\( Na + H_2O \rightarrow NaOH + H_2 \)[/tex]
A. [tex]\( Na + H_2O \rightarrow NaOH + H_2 \)[/tex]
In this equation, [tex]\( H_2O \)[/tex] is clearly present on the left-hand side of the reaction, which means it is a reactant.
B. [tex]\( H_2 + O_2 \rightarrow H_2O \)[/tex]
In this equation, [tex]\( H_2O \)[/tex] is found on the right-hand side, which means it is a product, not a reactant.
C. [tex]\( CO_2 + O_2 + \text{energy} \rightarrow C_6H_{12}O_6 + O_2 \)[/tex]
Here, [tex]\( H_2O \)[/tex] does not appear on either side of the equation, meaning it is neither a reactant nor a product.
D. [tex]\( Na + Cl \rightarrow NaCl \)[/tex]
In this equation, [tex]\( H_2O \)[/tex] is also not present on either side of the reaction.
Based on this analysis, the equation that has water (H₂O) as a reactant is:
A. [tex]\( Na + H_2O \rightarrow NaOH + H_2 \)[/tex]