<u>Answer:</u> The number of moles of carbon dioxide produced are
and the mass of calcium carbonate is 61.6 mg
<u>Explanation:</u>
To calculate the moles of carbon dioxide gas, we use the equation given by ideal gas which follows:

where,
P = pressure of the gas = 738.0 mmHg
V = Volume of the gas = 15.5 mL = 0.0155 L (Conversion factor: 1 L = 1000 mL)
T = Temperature of the gas = ![25^oC=[25+273]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B25%2B273%5DK%3D298K)
R = Gas constant = 
n = number of moles of carbon dioxide gas = ?
Putting values in above equation, we get:

For the given chemical equation:

By Stoichiometry of the reaction:
1 mole of carbon dioxide gas is produced from 1 mole of calcium carbonate
So,
of carbon dioxide will be produced from =
of calcium carbonate
- To calculate the mass for given number of moles, we use the equation:

Molar mass of calcium carbonate = 100 g/mol
Moles of calcium carbonate = 
Putting values in above equation:

Converting this into milligrams, we use the conversion factor:
1 g = 1000 mg
So, 
Hence, the number of moles of carbon dioxide produced are
and the mass of calcium carbonate is 61.6 mg