How many grams of CaCO3 will be produced if we produce 59.08 grams of NaCl in the reaction:

CaCl2 + Na2CO3 --> 2NaCl +CaCO3

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How many grams of CaCO3 will be produced if we produce 59.08 grams of NaCl in the reaction:

CaCl2 + Na2CO3 --> 2NaCl +CaCO3

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## Answers (2)

CaCl2(aq) + Na2CO3(aq) --> 2NaCl(aq) +CaCO3(s)

.................. ................ ....... 59.08g ...... ???g

Simply set up a series of conversion factors that take you from grams of NaCl to moles of NaCl to moles of CaCO3 to grams of CaCO3:

59.08 g NaCl x (1 mol NaCl / 58.44 g NaCl) x (1 mol CaCO3 / 2 mol NaCl) x (100.09 g CaCO3 / 1 mol CaCO3) = 50.59 g CaCO3

if you divide 59.08 by the Molecular Weight of NaCl you will get the moles of NaCl produced

if you then recognize from the equation that you will get one mole of CaCO3 for every two moles of NaCl produced, you divide the moles of NaCl by 2, and get the moles of CaCO3

then you multiply the moles of CaCO3 by the Molecular Weight of CaCO3 and you get the mass (in grams if you used molecular weights in the normal units of grams/mole).

makes sense?