Acid precipitation dripping on limestone produces carbon dioxide by the following reaction: caco3(s) + 2h+(aq) > ca(2+)(aq) + co2(g) + h2o (l) 15.5ml of co2 was produced at 25*c and 738.0 mmhg how man moles of co2 were produced? how many milligrams of caco3 were consumed?

The number of moles of carbon dioxide produced are and the mass of calcium carbonate is 61.6 mg

Explanation:

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 =

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

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

the answer is true.

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"