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Stoichiometric Calculations And Gas Volume Determinations In Chemical Reactions

This Solved Assignment covers stoichiometry and gas laws. Download now!

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Stoichiometric Calculations and Gas Volume Determinations in Chemical Reactions1)The balanced equation is:𝐿𝑖𝑂𝐻+π»π΅π‘Ÿβ†’πΏπ‘–π΅π‘Ÿ+𝐻2𝑂We have : n(LiOH)used= n(LiBr)produced= m(LiOH)/M(LiOH) =106.941+15.9994+1.00794=0.418π‘šπ‘œπ‘™Then m(LiBr) =M(LiBr) * n(LiBr) = (6.941+79.904)*0.418 = 36.3 g2)The balanced equation is:𝐢2𝐻4+3𝑂2β†’2𝐢𝑂2+2𝐻2𝑂We have : n(C2H4)used= n(CO2)produced/2then n(CO2)produced=2βˆ—454βˆ—1.00794+2βˆ—12.0107=3.208π‘šπ‘œπ‘™then m(CO2) = M(CO2) * n(CO2) = (12.0107+2*15.9994)*3.208 = 141.2 g3)The balanced equation is:𝑀𝑔+2π‘π‘ŽπΉβ†’π‘€π‘”πΉ2+2π‘π‘ŽWe have : n(Mg)used= n(Na)produced/2then n(Na)produced=2βˆ—5.524.305=0.453π‘šπ‘œπ‘™Therefore m(Na) = M(Na) * n(Na) = 22.989770 * 0.453 = 10.4 g4)The balanced equation is:2𝐻𝐢𝑙+π‘π‘Ž2𝑆𝑂4β†’2π‘π‘ŽπΆπ‘™+𝐻2𝑆𝑂4We have : n(HCl)used/2= n(H2SO4)produced=202βˆ—(1.00794+35.453)=0.274π‘šπ‘œπ‘™then m(H2SO4) =M(H2SO4) * n(H2SO4) = (2*1.00794 + 32.066 + 4*15.9994)*0.274 = 26.9 g1)The balanced equation is:𝑀𝑔+2𝐻𝐢𝑙→𝐻2+𝑀𝑔𝐢𝑙2We have : n(Mg)used= n(H2)produced=2024.305=0.823π‘šπ‘œπ‘™Then the number of liters of H2 gas produced is : 0.823 * 22.4 = 18.4 L2)The balanced equation is:2𝐾𝐢𝑙𝑂3β†’3𝑂2+2𝐾𝐢𝑙We have : n(KClO3)used/2 = n(O2)produced/3 then n(O2)produced=32βˆ—7.5=11.25π‘šπ‘œπ‘™Then the number of liters of H2 gas produced is :11.25* 22.4 =252L3)The balanced equation is:2π‘π‘ŽπΌ+𝐢𝑙2β†’2π‘π‘ŽπΆπ‘™+𝐼2We have : n(Cl2)used= n(NaCl)produced/2 then n(NaCl)produced=2βˆ—π‘›(𝐢𝑙2)=2βˆ—45.322.4=4.04π‘šπ‘œπ‘™4)The balanced equation is:2𝐾+2𝐻2𝑂→2𝐾𝑂𝐻+𝐻2We have : n(K)used/2 = n(H2)producedthen n(K)used=𝑛(𝐻2)2=(35.622.4)2=0.795π‘šπ‘œπ‘™Then m(K) = M(K) * n(K) = 39.0983 * 0.795 = 31.1 g

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