Question

hello I need help to calculate the corresponding stoichiometric mass in milligrams of CaCO3the answer to number 8 is 2.72 moles and I'll take a picture of equation 2

Answer 1

We have the balanced equation of the reaction, this is a neutralization reaction.

We have the moles of CO2 generated, equal to 2.72 moles. The number of moles of CaCO3 is found by reviewing the stoichiometric coefficients of equation 2. We have that the ratio CaCO3 to CO2 is 1/1.

Therefore, the moles of CaCO3 will be:

`molCaCO_3=2.72molCO_2*(1molCaCO_3)/(1molCO_2)=2.72molCaCO_3`

The mass of CaCO3 we will be found by multiplying the moles of CaCO3 by its molar mass. The molar mass of CaCO3 is 100.0869g/mol.

`\begin{gathered} MassCaCO_3=2.72molCaCO_3*(MolarMass,gCaCO_3)/(1molCaCO_3) \\ MassCaCO_3=2.72molCaCO_3*(100.0869gCaCO_3)/(1molCaCO_3)=272.24gCaCO_3 \\ MassCaCO_3=2.72*10^5mgCaCO_3 \end{gathered}`

Answer: The mass of CaCO3 needed will be 2.72x10^5 milligrams