Stoichiometry of neutralization:\nWhen 56 g of CaO are mixed with 63 g of HNO3, how many grams of Ca(NO3)2 are formed?

Introduction / Context:
Accurate stoichiometric calculations underpin reactor sizing and yield estimation. This problem involves the neutralization of a basic oxide with nitric acid to form a salt and water, emphasizing limiting-reagent identification and mass–mole conversions.

Given Data / Assumptions:

• Chemical reaction: CaO + 2 HNO3 → Ca(NO3)2 + H2O.
• Molar masses: CaO = 56 g/mol; HNO3 = 63 g/mol; Ca(NO3)2 = 164 g/mol.
• Masses provided: 56 g CaO and 63 g HNO3.

Concept / Approach:
Convert masses to moles, identify the limiting reactant using stoichiometric coefficients, then compute product moles from the limiting reagent. Finally, convert moles of product to mass using its molar mass.

Step-by-Step Solution:
n_CaO = 56 g / 56 g/mol = 1.0 mol.n_HNO3 = 63 g / 63 g/mol = 1.0 mol.Stoichiometry requires 2 mol HNO3 per 1 mol CaO; available HNO3 (1.0 mol) is limiting.From 2 HNO3 → 1 Ca(NO3)2, so 1.0 mol HNO3 yields 0.5 mol Ca(NO3)2.Mass of product = 0.5 mol × 164 g/mol = 82 g.

Verification / Alternative check:
If HNO3 were doubled (126 g), CaO would then be limiting, and 1 mol CaO would give 1 mol Ca(NO3)2 = 164 g, confirming the proportionality of our 82 g result.

Why Other Options Are Wrong:
164 g assumes excess acid; 41 g and 8.2 g are fractional values inconsistent with stoichiometry; 123 g lacks any stoichiometric basis.

Common Pitfalls:
Forgetting the 2:1 acid-to-oxide stoichiometry; using atomic instead of formula molar masses.

Final Answer:
82 g