Calcium nitrate and ammoonium fluoride react to form calcium fluoride, dinitrogenmonoxide and water vapor. What mass of each substance is present after 28.42 g of calcium nitrate and 29.60 g of ammonium fluoride react completely?

Answer 1

You have a limiting reagent problem when both reagents are listed with grams or moles.
1. Write the equation and balance it.
2a. Calculate moles Ca(NO3)2. moles = g/molar mass.
2b. Calculate moles NH4F.

3a. Convert moles of one of the reactants to moles of the other.
3b. Determine the limiting reagent.
3c. ALL of the limiting reagent will be used and there will be zero left.
3d. Subtract moles of the other reactant from initial moles.
3e. Calculate grams of the other reactant remaining.
Post your work if you get stuck.

Answer 2

28.42g/164.1= .17318 mol Ca(NO3)2

29.6g/37.042=.79909 mol (NH4)F

(1mol Ca(NO3)2 /.17318) x (2mol (NH4)F/x)=.34636

(2 mol (NH4)F/.79909)x(1 mol Ca(NO3)2/x)=.399545

so (NH4)F (.34636) is limiting reagent.

now I'm stuck. Not sure what you mean about subtracting moles from initial moles etc.....thanks!