Stoichiometric Calculations and Gas Volume Determinations in Chemical Reactions1)The balanced equation is:πΏπππ»+π»π΅πβπΏππ΅π+π»2πWe have : n(LiOH)used= n(LiBr)produced= m(LiOH)/M(LiOH) =106.941+15.9994+1.00794=0.418πππThen m(LiBr) =M(LiBr) * n(LiBr) = (6.941+79.904)*0.418 = 36.3 g2)The balanced equation is:πΆ2π»4+3π2β2πΆπ2+2π»2πWe have : n(C2H4)used= n(CO2)produced/2then n(CO2)produced=2β454β1.00794+2β12.0107=3.208πππthen m(CO2) = M(CO2) * n(CO2) = (12.0107+2*15.9994)*3.208 = 141.2 g3)The balanced equation is:ππ+2πππΉβπππΉ2+2ππWe have : n(Mg)used= n(Na)produced/2then n(Na)produced=2β5.524.305=0.453πππTherefore m(Na) = M(Na) * n(Na) = 22.989770 * 0.453 = 10.4 g4)The balanced equation is:2π»πΆπ+ππ2ππ4β2πππΆπ+π»2ππ4We have : n(HCl)used/2= n(H2SO4)produced=202β(1.00794+35.453)=0.274πππthen m(H2SO4) =M(H2SO4) * n(H2SO4) = (2*1.00794 + 32.066 + 4*15.9994)*0.274 = 26.9 g1)The balanced equation is:ππ+2π»πΆπβπ»2+πππΆπ2We have : n(Mg)used= n(H2)produced=2024.305=0.823πππThen the number of liters of H2 gas produced is : 0.823 * 22.4 = 18.4 L2)The balanced equation is:2πΎπΆππ3β3π2+2πΎπΆπWe have : n(KClO3)used/2 = n(O2)produced/3 then n(O2)produced=32β7.5=11.25πππThen the number of liters of H2 gas produced is :11.25* 22.4 =252L3)The balanced equation is:2πππΌ+πΆπ2β2πππΆπ+πΌ2We have : n(Cl2)used= n(NaCl)produced/2 then n(NaCl)produced=2βπ(πΆπ2)=2β45.322.4=4.04πππ4)The balanced equation is:2πΎ+2π»2πβ2πΎππ»+π»2We have : n(K)used/2 = n(H2)producedthen n(K)used=π(π»2)2=(35.622.4)2=0.795πππThen m(K) = M(K) * n(K) = 39.0983 * 0.795 = 31.1 gStoichiometric Calculations And Gas Volume Determinations In Chemical Reactions
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