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<span>(g). m(HgO) = 4.37 g. n</span>(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) = <span>0.02 mol; amount of substance. m</span>(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%. <span>yield = 61.77%.
3. Answer is: 68.16 </span><span>grams of the excess reactant (oxygen) remain. </span>Balanced chemical reaction: 4Fe(s) + 3O₂(g)→ 2Fe₂O₃<span>(g). m(Fe) = 27.3 g. n</span>(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. <span>Empirical
formula gives the proportions of the elements present in a compound. </span>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 (<span>number of particles (ions,</span> atoms<span> or </span>molecules), that are contained in <span>one </span>mole of substance<span>). </span> 7. Answer is: iron (Fe) <span>is the limiting reactant. </span>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.<span>
N(C₆H₁₄) = 2.62·10²³; number of molecules.</span><span>
n(C₆H₁₄) = N(C₆H₁₄) ÷ Na.</span><span>
n(C₆H₁₄) = 2.62·10²³ ÷ 6.022·10²³ 1/mol.</span><span>
n(C₆H₁₄) = 0.435 mol; amount of substance of </span>C₆H₁₄.<span>
Na - Avogadro constant or Avogadro number. </span> 9. Answer is: 3.675 <span>moles of carbon(II) oxide are required to completely react. </span>Balanced chemical reaction: Fe₂O₃<span>(s) + 3CO(g) ⟶ 2Fe(s) + 3CO</span>₂<span>(g). n(</span>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.<span>
N(Ba</span><span>) = 2.62·10²³; number of
atoms of barium.
n</span>(Ba) = N(Ba)<span> ÷ Na.
n</span>(Ba) = 1.3·10²⁴<span> ÷
6.022·10²³ 1/mol.
n</span>(Ba)<span> = 2.158 mol; amount of
substance of barium</span>.<span>
Na - Avogadro constant or Avogadro number.
11. Answer is: </span>6.26·10²³ <span>carbon atoms are present. </span>n(C₂H₆O) = 0.52 mol; amount of substance.<span>
N</span>(C₂H₆O) =
n(C₂H₆O) · Na.<span>
N</span>(C₂H₆O) =
0.52 mol · 6.022·10²³ 1/mol.<span>
N</span>(C₂H₆O) =
3.13·10²³.<span>
In one molecule of </span>C₂H₆O there are two atoms of carbon:<span> N(C</span>) = N(C₂H₆O) · 2. N(C) = 3.13·10²³ · 2. N(C) = 6.26·10²³. <span> 12. Answer is: </span><span>the empirical formula is C</span>₂H₄O.<span> </span><span>If we use 100 grams of compound: </span>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.
