2c2h2(l) 5o2(g) yeild 4co2(g) 2h2o(g) if the acetylene tank contains 37.0 mol of c2h2 and the oxygen tank contains 81.0 mol of o2, what is the limiting reactant for this reaction?

2C2H2(l)  +5O2(g)→ 4Co2(g)  + 2H2O

The   limiting  reactant  for reaction above is O2

explanation

 The limiting reagent is determined using mole ratio  of both reactant.

 that  is the  mole ratio  of  C2H2:O2  which is  2:5 .

 The mole ratio above  implies  that 37.0 mole  of C2H2   needs  37.0 moles  x5/2=92  moles of  O2.

Since the available moles of O2  was 81.0 mole  and 92  moles  are  required to completely  react with  C2H2 , O2   is  the limiting reagent.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%