Question

Part A Calcium carbonate (CaCOs) reacts with stomach acid (HCI hydrochionic acid) according to the following equation Atypical antacid contains CaCO, ll sch an antacid is added to 100mLofasou ton that s ? 300 Mnila how many ga satch gas ae poacen CaCO, (s) + 2HCl(aq)-+ CO,(g) + F,O() + CaCh(aq) Express the mass to three significant figures and include the appropriate units

Answer 1

The balanced equation is given as:

CaCO₃ (s) + 2HCl (aq) ---> CO₂ (g) + H₂O (l) + CaCl₂ (aq)

To calculate the amount of CO₂ gas produced in this reaction, we follow certain steps:

Step 1: Calculation of No. of moles of HCl

The number of moles of HCl will be = (10.0/1000) L X 0.300 M = 0.003 mole [M = mole/L]

(Since, Molarity = No. of moles of solute / Volume of solution in liters)

Step 2: Calculation of moles of CO₂ from moles of HCl using the balanced chemical equation

2 moles of HCl produces = 1 mole of CO₂

Hence, 0.003 mole of HCl will produce = (1/2) X 0.003 mole = 1.5 X 10^-3 mole of CO₂

Step 3: Convert the number of moles of CO₂ to grams of CO₂

Grams of CO₂ gas produced are = 1.5 X 10^-3 mole X 44 g/mol   [Molar mass of CO₂ = 44 g/mol]

= 0.0660 g

(Since, Grams of gas produced = No. of moles of gas X Molar mass of gas)

Therefore, 0.0060 g of CO₂ gas are produced in this reaction.