What volume (in mLmL) of a 0.175 MHNO3MHNO3 solution is required to completely react with 44.2 mLmL of a 0.108 MNa2CO3MNa2CO3 solution according to the following balanced chemical equation? Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2 (g)+H2O(l)

54.5 mL

Explanation:

Let's consider the following balanced equation.

Na₂CO₃(aq) + 2 HNO₃(aq) → 2 NaNO₃(aq) + CO₂(g) + H₂O(l)

44.2 mL of 0.108 M Na₂CO₃ solution react. The moles of Na₂CO₃ that react are:

0.0442 L × 0.108 mol/L = 4.77 × 10⁻³ mol

The molar ratio of Na₂CO₃ to HNO₃ is 1:2. The moles of HNO₃ that reacted are  2/1 × 4.77 × 10⁻³ mol = 9.54 × 10⁻³ mol

9.54 × 10⁻³ moles of HNO₃ are in a 0.175 M solution. The required volume of HNO₃ is:

9.54 × 10⁻³ mol × (1 L/0.175 mol) × (1000 mL/1 L) = 54.5 mL

your answer would be “dry”