Question

85 POINTS PLEASE HELP ASAP

Answer 1

1. Answer is: there are 1.41·10²³ molecules of oxygen.
1) calculate amount of substance for oxygen gas:
m(O₂) = 7.5 g; mass of oxygen.
n(O₂) = m(O₂) ÷ M(O₂).
n(O₂) = 7.5 g ÷ 32 g/mol.
n(O₂) = 0.234 mol.
2) calculate number of molecules:
N(O₂) = n(O₂) · Na.
N(O₂) = 0.234 mol · 6.022·10²³ 1/mol.
N(O₂) = 1.41·10²³.
Na - Avogadro constant.

2. Answer is: the percent yield for the reaction is 61.77%.
Balanced chemical reaction: 2HgO(s) → 2Hg(l) + O₂(g).
m(HgO) = 4.37 g.
n(HgO) = n(HgO) ÷ M(HgO).
n(HgO) = 4.37 g ÷ 216.6 g/mol.
n(HgO) = 0.02 mol.
From chemical reaction: n(HgO) : n(Hg) = 2 : 2 (1 :1).
n(Hg) = n(HgO) = 0.02 mol; amount of substance.
m(Hg) = n(Hg) ·M(Hg).
m(Hg) = 0.02 mol · 200.6 g/mol.
m(Hg) = 4.047 g.
yield = 2.5 g ÷ 4.047 g · 100%.
yield = 61.77%.

3. Answer is: 68.16  grams of the excess reactant (oxygen) remain.
Balanced chemical reaction: 4Fe(s) + 3O₂(g)→ 2Fe₂O₃(g).
m(Fe) = 27.3 g.
n(Fe) = m(Fe) ÷ M(Fe).
n(Fe) = 27.3 g ÷ 55.85 g/mol.
n(Fe) = 0.489 mol.
m(O₂) = 79.9 g.
n(O₂) = 79.9 g ÷ 32 g/mol.
n(O₂) = 2.497 mol; amount of substance.
From chemical reaction: n(Fe) . n(O₂) = 4 : 3.
0.489 mol : n(O₂) = 4 : 3.
n(O₂) = 3 · 0.489 mol ÷ 4.
n(O₂) = 0.367 mol.
Δn(O₂) = 2.497 mol - 0.367 mol.
Δn(O₂) = 2.13 mol.
m(O₂) = 2.13 mol · 32 g/mol.
m(O₂) = 68.16 g.

4. Answer is: there are 0.603 moles of ammonia.
m(NH₃) = 10.25 g; mass of ammonia.
M(NH₃) = Ar(N) + 3Ar(H) · g/mol.
M(NH₃) = 14 + 3·1 · g/mol.
M(NH₃) = 17 g/mol; molar mass of ammonia.
n(NH₃) = m(NH₃) ÷ M(NH₃).
n(NH₃) = 10.25 g ÷ 17 g/mol.
n(NH₃) = 0.603 mol; amount of substance (ammonia).

5. Answer is: the empirical formula mass of P₂O₅ is 141.89.
Empirical formula gives the proportions of the elements present in a compound.
Atomic mass of phosphorus is 30.97 g/mol.
Atomic mass of oxygen is 15.99 g/mol.
In phosphorus (V) oxide there are two atoms of phosphorus and five atoms of oxygen:
EFM(P₂O₅) = 2·30.97 g/mol + 5·15.99 g/mol = 141.89 g/mol. 

6. Answer is: there are 1.108·10²⁴ molecules of water.
n(H₂O) = 1.84 mol; amount of substance (water).
N(H₂O) = n(H₂O) · Na.
N(H₂O) = 1.84 mol · 6.022·10²³ 1/mol.
N(H₂O) = 11.08·10²³.
N(H₂O) = 1.108·10²⁴.
Na - Avogadro constant (number of particles (ions, atoms or molecules), that are contained in one mole of substance).

7. Answer is: iron (Fe) is the limiting reactant.
Balanced chemical reaction: 4Fe(s) + 3O₂(g)→ 2Fe₂O₃(g).
m(Fe) = 27.3 g.
n(Fe) = m(Fe) ÷ M(Fe).
n(Fe) = 27.3 g ÷ 55.85 g/mol.
n(Fe) = 0.489 mol.
m(O₂) = 45.8 g.
n(O₂) = 45.8 g ÷ 32 g/mol.
n(O₂) = 1.431 mol; amount of substance.
From chemical reaction: n(Fe) . n(O₂) = 4 : 3.
For 1.431moles of oxygen we need:
1.431 mol : n(Fe) = 3 : 4.
n(Fe) = 1.908 mol, there is no enough iron.

8. Answer is: there are 0.435 moles of C₆H₁₄.
N(C₆H₁₄) = 2.62·10²³; number of molecules.
n(C₆H₁₄) = N(C₆H₁₄) ÷ Na.
n(C₆H₁₄) = 2.62·10²³ ÷ 6.022·10²³ 1/mol.
n(C₆H₁₄) = 0.435 mol; amount of substance of C₆H₁₄.
Na - Avogadro constant or Avogadro number.

9. Answer is: 3.675 moles of carbon(II) oxide are required to completely react.
Balanced chemical reaction: Fe₂O₃(s) + 3CO(g) ⟶ 2Fe(s) + 3CO₂(g).
n(Fe₂O₃) = 1.225 mol; amount of substance.
From chemical reaction: n(Fe₂O₃) : n(CO) = 1 : 3.
1.225 mol : n(CO) = 1 : 3.
n(CO) = 3 · 1.225 mol.
n(CO) = 3.675 mol.

10. Answer is: there are 2.158 moles of barium atoms.
N(Ba) = 2.62·10²³; number of atoms of barium.
n(Ba) = N(Ba) ÷ Na.
n(Ba) = 1.3·10²⁴ ÷ 6.022·10²³ 1/mol.
n(Ba) = 2.158 mol; amount of substance of barium.
Na - Avogadro constant or Avogadro number.

11. Answer is: 6.26·10²³ carbon atoms are present.
n(C₂H₆O) = 0.52 mol; amount of substance.
N(C₂H₆O) = n(C₂H₆O) · Na.
N(C₂H₆O) = 0.52 mol · 6.022·10²³ 1/mol.
N(C₂H₆O) = 3.13·10²³.
In one molecule of C₂H₆O there are two atoms of carbon:
N(C) = N(C₂H₆O) · 2.
N(C) = 3.13·10²³ · 2.
N(C) = 6.26·10²³.

12. Answer is: the empirical formula is C₂H₄O.
If we use 100 grams of compound:
1) ω(C) = 51% ÷ 100% = 0.51.
m(C) = ω(C) · m(compound).
m(C) = 0.51 · 100 g.
m(C) = 51 g.
n(C) = m(C) ÷ M(C).
n(C) = 51 g ÷ 12 g/mol.
n(C) = 4.25 mol.
2) ω(H) = 9.3 % ÷ 100% = 0.093.
m(H) = 0.093 · 100 g.
m(H) = 9.3 g.
n(H) = 9.3 g ÷ 1 g/mol.
n(H) = 9.3 mol
3) ω(O) = 39.2 % ÷ 100%.
ω(O) = 0.392.
m(O) = 0.392 · 100 g.
m(O) = 39.2 g.
n(O) = 39.2 g ÷ 16 g/mol.
n(O) = 2.45 mol.
4) n(C) : n(H) : n(O) = 4.25 mol : 9.3 mol : 2.45 mol /2.45 mol.
n(C) : n(H) : n(O) = 1.73 : 3.795 : 1.