For the reaction, calculate how many grams of the product form when 3.2 g of CaO completely reacts.
Assume that there is more than enough of the other reactant.
CaO(s)+CO2(g)→CaCO3(s)
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To calculate the grams of product formed, you need to use stoichiometry and the balanced chemical equation. Here's how you can do it step by step:
Step 1: Write the balanced chemical equation:
CaO(s) + CO2(g) → CaCO3(s)
Step 2: Calculate the molar mass of CaO:
CaO: 40.08 g/mol (calcium) + 16.00 g/mol (oxygen) = 56.08 g/mol
Step 3: Use the molar mass of CaO to convert grams to moles:
Moles of CaO = grams / molar mass of CaO
Moles of CaO = 3.2 g / 56.08 g/mol = 0.057 mol
Step 4: Use the stoichiometry of the balanced chemical equation to find the moles of the product formed.
From the balanced equation, the ratio of CaO to CaCO3 is 1:1, meaning that for every 1 mole of CaO, you get 1 mole of CaCO3.
Moles of CaCO3 = Moles of CaO (based on stoichiometry)
Moles of CaCO3 = 0.057 mol
Step 5: Convert moles of CaCO3 to grams using the molar mass of CaCO3:
Molar mass of CaCO3: 40.08 g/mol (calcium) + 12.01 g/mol (carbon) + 3(16.00 g/mol) (oxygen) = 100.09 g/mol
Grams of CaCO3 = Moles of CaCO3 × molar mass of CaCO3
Grams of CaCO3 = 0.057 mol × 100.09 g/mol = 5.71 g
Therefore, 3.2 g of CaO will completely react to form 5.71 g of CaCO3.