0.349g of cacl2*2h2o and 0.698g of na2co3 are dissolved in 100ml of water to form a solution.?
ID: 819085 • Letter: 0
Question
Which is the limiting reactant?How many grams of CaCO3 will precipitate?
How many grams of excess reactant remain after the reaction has gone tell completion?
I know the overall equation is:
Na2CO3(aq)+CaCl2*H2O(aq)--->CaCO3(s)+2N...
But I'm confused on how to get the rest. Which is the limiting reactant?
How many grams of CaCO3 will precipitate?
How many grams of excess reactant remain after the reaction has gone tell completion?
I know the overall equation is:
Na2CO3(aq)+CaCl2*H2O(aq)--->CaCO3(s)+2N...
But I'm confused on how to get the rest.
Explanation / Answer
(1) CaCl2.2H2O + Na2CO3 => CaCO3 + 2 NaCl + 2 H2O
Theoretical moles of CaCl2 : Na2CO3 = 1 : 1
Actual moles of CaCl2.2H2O = mass/molar mass of CaCl2.2H2O = 0.349/147.015 = 0.002374 mol
Actual moles of Na2CO3 = mass/molar mass of Na2CO3 = 0.698/105.99 = 0.006586 mol
Actual moles of CaCl2.2H2O : Na2CO3 = 0.002374 : 0.006586 = 1 : 2.774
Since Na2CO3 is in excess => CaCl2.2H2O is the limiting reactant
(2) Moles of CaCO3 = moles of CaCl2.2H2O = 0.002374 mol
Mass of CaCO3 = moles x molar mass of CaCO3
= 0.002374 x 100.0875
= 0.238 g
(3) Moles of Na2CO3 reacted = moles of CaCl2.2H2O = 0.002374 mol
Mass of Na2CO3 reacted = moles x molar mass of Na2CO3
= 0.002374 x 105.99 = 0.252 g
Mass of Na2CO3 remaining = 0.698 - 0.252
= 0.446 g
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