If 75.2 grams of Calcium Sulfate are reacted with an unlimited amount of Sodium Phosphate,
what mass of Calcium Phosphate will be formed according to the reaction:

55.83 grams is the mass of Calcium Phosphate will be formed according to the reaction when 75.2 grams of Calcium Sulfate are reacted with an unlimited amount of Sodium Phosphate.

Explanation:

Data given:

mass of calcium sulphate reacted= 75.2 grams

atomic mass of calcium sulphate = 136.14 grams/mole

balanced chemical reaction equation:

3CaS +2 (P)   ⇒   (P) + 3S

number of moles of calcium phosphate given will be calculated as:

number of moles =

        putting the values in the above formula, we get

number of moles =

                             = 0.55 moles of CaS is given

From the balanced equation:

3 moles of CaS reacts to give 1 mole of calcium phosphate

so, 0.55 moles of CaS will react to form x moles of calcium phosphate

=

3x = 0.55

x = 0.18 moles of calcium phosphate is formed

mass = number of moles x atomic mass of calcium phosphate

(atomic mass of (P) =310.18 grams/mole)

putting the values in the formula:

mass = 0.18 x 310.18

          = 55.83 grams of calcium phosphate is formed.

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(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied