Balance the following chemical reaction:
Na3P + CaF2 --> NaF +
+
Ca3P2

Let a, b, c, d be constants such that

a Na₃P + b CaF₂   →   c NaF + d Ca₃P₂

Count the atoms on either side.

Left:

Na = 3a, P = a, Ca = b, F = 2b

Right:

Na = c, F = c, Ca = 3d, P = 2d

To get the same number of atoms of each element on both sides, we would have

3a = c

a = 2d

b = 3d

2b = c

If we let d = 1, we get a = 2, b = 3, and c = 6, so the balanced equation is

2 Na₃P + 3 CaF₂   →   6 NaF + Ca₃P₂

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂

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