Question

Consider the following reaction. How many moles of oxygen are required to produce 8.00 moles of water? Assume that there is excess C,H,SH present. CH,SH() + 6 O2(g) + 3 CO2(g) + O2(g) + 4H2O(g) 8.00 moles 02 12.0 moles 02 2.00 moles Oz 5.33 moles 02 32.0 moles 02 Submit Request Provide Feedback

Answer 1

1) Answer :- 12 moles O₂.

The given reaction is

C3H7SHU) + 602(g) + 3CO2(g) + SO2(g) + 4H2O(g)

Form reaction it is clear that,

4 moles of water produced from 6 moles of oxygen.

ie. 4 mol of water = 6 mol of oxygen

therefore,

8 mol of water =
8(mol) x 6(mol) = 12(mol 4(mol)
of oxygen.

therefore, 12 mol of oxygen will require to produce 8 mol of water.

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2) Answer :- 108 g NaCl

The given balanced reaction is

2Na(s) + Cl2(g) + 2Nacho
  

from reaction

theoretical Ratio of moles of reactants, Na:Cl₂ = $\frac{2}{1} = 1$

Given :-

Mass of Na = 58.8 g

Molar mass of Na = 23 g/mol

Mass of Cl₂ = 65.5 g

Molar mass of Cl₂ = 71 g/mol

we know
Mass Moles = Molar Mass

therefore,

Moles of Na =
58.8 (9) = 2.56mol 23(g/mol)

Moles of Cl₂ =
65.5 (9) - = 0.923mol 71(g/mol)

therefore

Ratio of actual moles of Na:Cl₂ is
2.56 0.925=277

ie. theoretical mol ratio < actual mol ratio.

Hence, Cl₂ is limiting reagent.

Thus theoretical yield is calculated from Cl₂

from reaction

1 mol of Cl₂ = 2 mol of NaCl

therefore,

0.923 mol of Cl₂ =
0.923(mol) x 2(mol) 2 = 1.846(mol) 1(mol)
of NaCl.

Now,

Molar mass of NaCl = 58.5 g/mol.

therefore,

Mass of NaCl =
1.846(mol) x 58.5(g/mol) = 107.9(9) = 108(9)