Consider the reaction cacn2 + 3 h2o → caco3 + 2 nh3 . how much caco3 is produced if 47.5 moles nh3 are produced? 1. 0.950 g 2. 9500 g 3. 0.238 g 4. 23.8 g 5. 4750 g 6. 2380 g

The  CaCO3  produced  if  47.5   moles of  NH3  produced  is   calculated  as  follows

CaCN2  +3H2O  = CaCO3  + 2NH3

by  use   of  mole  ratio  between  CaCO3 to NH3  which  is 1:2  the  moles  of  CaCO3  is therefore =  47.5 /2= 23.75  moles

mass of CaCO3   is  therefore = moles  x  molar  mass
= 23.75 moles  x  100g/mol=  2375 grams   which  is approximate   2380  grams(answer 6)

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