Stoichiometry And Stoichiometric Calculations

To determine the amount of HCl that reacts completely with 15 g of CaCO₃, we can use the stoichiometry of the reaction:

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

The first step is to calculate the molar mass of CaCO₃ and HCl:

• Molar mass of CaCO₃ = 40 (Ca) + 12 (C) + 3×16 (O) = 100 g/mol
• Molar mass of HCl = 1 (H) + 35.5 (Cl) = 36.5 g/mol

Next, find the moles of CaCO₃ in 15 g:

Moles of CaCO₃ = 15 g100 g/mol = 0.15 moles

According to the balanced equation, 1 mole of CaCO₃ reacts with 2 moles of HCl. Thus, 0.15 moles of CaCO₃ will require:

0.15 moles of CaCO₃ × 2 moles of HCl/mole of CaCO₃ = 0.30 moles of HCl

Now, convert moles of HCl to grams:

Mass of HCl = 0.30 moles × 36.5 g/mol = 10.95 g

Therefore, 15 g of CaCO₃ completely reacts with 10.95 g of HCl.

From the reaction: We are given the rate of formation of Z as .
We need to find the rate of disappearance of X. The stoichiometric relationship between X and Z is given by the coefficient ratio.
For every 3 moles of Z formed, 5 moles of X are consumed. Thus, the rate of disappearance of X can be related to the rate of formation of Z by the following equation: Substituting the given rate of formation of Z: Thus, the correct answer is .