Question 8.11: Calcium carbonate, CaCO3, in antacids reacts with HCl in the......

STEP 1   State the given and needed quantities.
Table 1
STEP 2   Write a plan to convert the given quantity to the needed moles.
grams  of  CaCO_{3}    \boxed{Molar  Mass} →      moles  of  CaCO_{3}    \boxed{Mole-mole  factor } →    moles  of  CO_{2}​
STEP 3   Write the equalities and conversion factors for molar mass and mole–mole factors.
1 mole of CaCO_{3} = 100.09 g of CaCO_{3}    1 mole of CaCO_{3} = 1 mole of CO_{2}
\frac{100.09 \space g \space CaCO_{3}}{1 \space mole \space CaCO_{3}}   and   \frac{1 \space mole \space CaCO_{3}}{100.09 \space g \space CaCO_{3}}                 \frac{1 \space mole \space CaCO_{3}}{1 \space mole \space CO_{2}}   and   \frac{1 \space mole \space CO_{2}}{1 \space mole \space CaCO_{3}}
STEP 4   Set up the problem to calculate moles of needed quantity.
25.0 \space \cancel {g \space CaCO_{3}} \times \frac{1 \space \cancel{mole \space CaCO_{3}}}{100.09 \space \cancel {g \space CaCO_{3}}} \times \frac{1 \space mole \space CO_{2}}{1 \space \cancel{mole \space CaCO_{3}}} = 0.250 \space mole \space of \space CO_{2}
STEP 5   Convert the moles of needed quantity to volume using the ideal gas law equation.
V =\frac{nRT}{P}
V =\frac{0.250 \space \cancel{mole} \space \times \space \frac{62.4 \space L  ·  \cancel{mmHg}}{\cancel{mole}  ·  \cancel K} \space \times  \space 297 \space \cancel K}{752 \space \cancel {mmHg}}=6.16 \space L \space of \space CO_{2}