Given the following equation: Al2(SO3)3 + 6 NaOH → 3 Na2SO3 + 2 Al(OH)3 If 10.0 g of Al2(SO3)3 is reacted with 10.0 g of NaOH, determine the limiting and excess
reactants. How much Na2SO3 is produced? How much excess reactant is left over?

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120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890

(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