Stoichiometric Calculations And Gas Volume Determinations In Chemical Reactions
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Stoichiometric Calculations and Gas Volume Determinations in Chemical Reactions
1) The balanced equation is : πΏπππ» + π»π΅π β πΏππ΅π + π»2π
We have : n(LiOH)used = n(LiBr)produced = m(LiOH)/M(LiOH) = 10
6.941+15.9994+1.00794 = 0.418 πππ
Then m(LiBr) = M(LiBr) * n(LiBr) = (6.941+79.904)*0.418 = 36.3 g
2) The balanced equation is : πΆ2π»4 + 3π2 β 2πΆπ2 + 2π»2π
We have : n(C2H4)used = n(CO2)produced/2 then n(CO2)produced = 2 β 45
4β1.00794+2β12.0107 =
3.208 πππ then m(CO2) = M(CO2) * n(CO2) = (12.0107+2*15.9994)*3.208 = 141.2 g
3) The balanced equation is : ππ + 2πππΉ β πππΉ2 + 2ππ
We have : n(Mg)used = n(Na)produced/2 then n(Na)produced = 2 β 5.5
24.305 = 0.453 πππ
Therefore m(Na) = M(Na) * n(Na) = 22.989770 * 0.453 = 10.4 g
4) The balanced equation is : 2π»πΆπ + ππ2ππ4 β 2πππΆπ + π»2ππ4
We have : n(HCl)used/2 = n(H2SO4)produced = 20
2β(1.00794+35.453) = 0.274 πππ then m(H2SO4) =
M(H2SO4) * n(H2SO4) = (2*1.00794 + 32.066 + 4*15.9994)*0.274 = 26.9 g
1) The balanced equation is : ππ + 2π»πΆπ β π»2 + πππΆπ2
We have : n(Mg)used = n(H2)produced = 20
24.305 = 0.823 πππ
Then the number of liters of H2 gas produced is : 0.823 * 22.4 = 18.4 L
2) The balanced equation is : 2πΎπΆππ3 β 3π2 + 2πΎπΆπ
We have : n(KClO3)used/2 = n(O2)produced/3 then n(O2)produced = 3
2 β 7.5 = 11.25 πππ
Then the number of liters of H2 gas produced is : 11.25 * 22.4 = 252 L
3) The balanced equation is : 2πππΌ + πΆπ2 β 2πππΆπ + πΌ2
We have : n(Cl2)used = n(NaCl)produced/2 then n(NaCl)produced = 2 β π(πΆπ2) = 2 β 45.3
22.4 = 4.04 πππ
4) The balanced equation is : 2πΎ + 2π»2π β 2πΎππ» + π»2
We have : n(K)used/2 = n(H2)produced then n(K)used = π(π»2)
2 = (35.6
22.4)
2 = 0.795 πππ
Then m(K) = M(K) * n(K) = 39.0983 * 0.795 = 31.1 g
1) The balanced equation is : πΏπππ» + π»π΅π β πΏππ΅π + π»2π
We have : n(LiOH)used = n(LiBr)produced = m(LiOH)/M(LiOH) = 10
6.941+15.9994+1.00794 = 0.418 πππ
Then m(LiBr) = M(LiBr) * n(LiBr) = (6.941+79.904)*0.418 = 36.3 g
2) The balanced equation is : πΆ2π»4 + 3π2 β 2πΆπ2 + 2π»2π
We have : n(C2H4)used = n(CO2)produced/2 then n(CO2)produced = 2 β 45
4β1.00794+2β12.0107 =
3.208 πππ then m(CO2) = M(CO2) * n(CO2) = (12.0107+2*15.9994)*3.208 = 141.2 g
3) The balanced equation is : ππ + 2πππΉ β πππΉ2 + 2ππ
We have : n(Mg)used = n(Na)produced/2 then n(Na)produced = 2 β 5.5
24.305 = 0.453 πππ
Therefore m(Na) = M(Na) * n(Na) = 22.989770 * 0.453 = 10.4 g
4) The balanced equation is : 2π»πΆπ + ππ2ππ4 β 2πππΆπ + π»2ππ4
We have : n(HCl)used/2 = n(H2SO4)produced = 20
2β(1.00794+35.453) = 0.274 πππ then m(H2SO4) =
M(H2SO4) * n(H2SO4) = (2*1.00794 + 32.066 + 4*15.9994)*0.274 = 26.9 g
1) The balanced equation is : ππ + 2π»πΆπ β π»2 + πππΆπ2
We have : n(Mg)used = n(H2)produced = 20
24.305 = 0.823 πππ
Then the number of liters of H2 gas produced is : 0.823 * 22.4 = 18.4 L
2) The balanced equation is : 2πΎπΆππ3 β 3π2 + 2πΎπΆπ
We have : n(KClO3)used/2 = n(O2)produced/3 then n(O2)produced = 3
2 β 7.5 = 11.25 πππ
Then the number of liters of H2 gas produced is : 11.25 * 22.4 = 252 L
3) The balanced equation is : 2πππΌ + πΆπ2 β 2πππΆπ + πΌ2
We have : n(Cl2)used = n(NaCl)produced/2 then n(NaCl)produced = 2 β π(πΆπ2) = 2 β 45.3
22.4 = 4.04 πππ
4) The balanced equation is : 2πΎ + 2π»2π β 2πΎππ» + π»2
We have : n(K)used/2 = n(H2)produced then n(K)used = π(π»2)
2 = (35.6
22.4)
2 = 0.795 πππ
Then m(K) = M(K) * n(K) = 39.0983 * 0.795 = 31.1 g
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Subject
Chemistry