Answer:
The balanced formula for the given reaction is:
CaCl2 + Na2CO3 ⇋ CaCO3 + 2NaCl
From the balanced equation, we can see that for every 1 mole of CaCl2, we need 1 mole of Na2CO3 to react.
To determine the amount of Na2CO3 needed, we first need to convert the given amount of CaCl2 to moles. The molar mass of CaCl2 is 111 g/mol, so 2.22 grams of CaCl2 is:
2.22 g CaCl2 x (1 mol CaCl2/111 g CaCl2) = 0.02 moles CaCl2
Since we need 1 mole of Na2CO3 for every mole of CaCl2, we need 0.02 moles of Na2CO3.
To convert moles of Na2CO3 to grams, we need to use the molar mass of Na2CO3, which is 106 g/mol:
0.02 moles Na2CO3 x (106 g Na2CO3/1 mol Na2CO3) = 2.12 grams Na2CO3
Therefore, we need 2.12 grams of Na2CO3 for the reaction.
To determine the amount of product (CaCO3) formed, we need to use stoichiometry to find the number of moles of CaCO3 formed. From the balanced equation, we can see that for every 1 mole of CaCl2, 1 mole of CaCO3 is formed.
So, 0.02 moles of CaCl2 will produce 0.02 moles of CaCO3.
The molar mass of CaCO3 is 100 g/mol, so the mass of CaCO3 formed is:
0.02 moles CaCO3 x (100 g CaCO3/1 mol CaCO3) = 2 grams CaCO3
Therefore, we should form 2 grams of CaCO3 as the product.
Explanation: :)