calcium carbonate + ethanoic acid reaction moles and weight

[observer1]

CaCO₃ + 2CH₃COOH ---> Ca(CH₃COO)₂ + H₂O + CO₂

If i add 2 moles of calcium carbonate and 2 moles of vinegar, how much of each of the result will i get in moles or grams?

[exchemist]

On 1/13/2023 at 1:29 PM, observer1 said:

CaCO₃ + 2CH₃COOH ---> Ca(CH₃COO)₂ + H₂O + CO₂

If i add 2 moles of calcium carbonate and 2 moles of vinegar, how much of each of the result will i get in moles or grams?

Show us how you are going about this, first.

[observer1]

nvm i found it anyways

[apprentus]

The balanced equation shows that 1 mole of calcium carbonate reacts with 2 moles of acetic acid (vinegar) to produce 1 mole of calcium acetate, 1 mole of water, and 1 mole of carbon dioxide.

Therefore, if you add 2 moles of calcium carbonate and 2 moles of vinegar, all of the vinegar will be used up since there is enough vinegar to react with both moles of calcium carbonate. The reaction will produce:

• 2 moles of calcium acetate (Ca(CH3COO)2)
• 2 moles of water (H2O)
• 2 moles of carbon dioxide (CO2)

To calculate the amount of each product in grams, you need to know the molar mass of each compound. The molar masses are:

• Calcium acetate (Ca(CH3COO)2) = 158.17 g/mol
• Water (H2O) = 18.02 g/mol
• Carbon dioxide (CO2) = 44.01 g/mol

So, the amount of each product in grams is:

• Calcium acetate: 2 moles x 158.17 g/mol = 316.34 g
• Water: 2 moles x 18.02 g/mol = 36.04 g
• Carbon dioxide: 2 moles x 44.01 g/mol = 88.02 g

Therefore, if you add 2 moles of calcium carbonate and 2 moles of vinegar, you will get 316.34 g of calcium acetate, 36.04 g of water, and 88.02 g of carbon dioxide.