When 20 mL of 0.80 M HCl is mixed with 30 mL of 0.40 M NaOH, what mass of water (in grams) is formed from the neutralization reaction?

• A. 0.32 g
• B. 0.64 g
• C. 0.48 g
• D. 0.80 g

Answer: (C)

Verified step by step guidance

1. Write the balanced chemical equation for the neutralization reaction: \(\mathrm{HCl + NaOH \rightarrow NaCl + H_2O}\).
2. Calculate the moles of HCl using the formula \(\text{moles} = M \times V\), where \(M\) is molarity and \(V\) is volume in liters. For HCl: \$0.80\,M \times 0.020\,L$.
3. Calculate the moles of NaOH similarly: \$0.40\,M \times 0.030\,L$.
4. Determine the limiting reactant by comparing the moles of HCl and NaOH, since the reaction ratio is 1:1.
5. Calculate the moles of water formed, which equals the moles of the limiting reactant, then convert moles of water to grams using the molar mass of water (\$18.02\,g/mol\() with the formula \)\text{mass} = \text{moles} \times \text{molar mass}$.