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일반화학/[03장] 화학량론778

NH3 2.94 g excess NaClO NCl3 gram NH3 2.94 g excess NaClO NCl3 gram If chlorine bleach is mixed with other cleaning products containing ammonia, the toxic gas NCl3(g) can form according to the equation: 3NaClO(aq) + NH3(aq) → 3NaOH(aq) + NCl3(g) When 2.94 g of NH3 reacts with an excess of NaClO according to the preceding reaction, how many grams of NCl3 are formed? > NH3의 몰질량 = 17.03 g/mol > NCl3의 몰질량 = 120.37 g/mol 질량비로 계산하면, (.. 2024. 4. 1.
Al 0.792 g excess H2SO4 H2 gas 0.0813 g percent Al 0.792 g excess H2SO4 H2 gas 0.0813 g percent Aluminum metal reacts with sulfuric acid to form hydrogen gas and aluminum sulfate. a) Write a balanced chemical equation for this reaction. b) Suppose that a 0.792 g sample of aluminum that contains impurities is reacted with excess sulfuric acid and 0.0813 g of H2 is collected. Assuming that none of the impurities reacts with sulfuric acid to pro.. 2024. 3. 31.
Si(s) + 2Cl2(g) → SiCl4(l) SiCl4 0.507 mol Cl2 mol Si(s) + 2Cl2(g) → SiCl4(l) SiCl4 0.507 mol Cl2 mol Silicon tetrachloride (SiCl4) can be prepared by heating Si in chlorine gas: Si(s) + 2Cl2(g) → SiCl4(l) In one reaction, 0.507 mole of SiCl4 is produced. How many moles of molecular chlorine were used in the reaction? Si(s) + 2Cl2(g) → SiCl4(l) Cl2 : SiCl4 = 2 : 1 계수비(= 몰수비) 이므로, 0.507 mol SiCl4 생성 시, 반응한(사용된) Cl2의 몰수를 계산하면, Cl2 : SiCl4 = 2 : 1 .. 2024. 3. 30.
1 g당 분자수 = 1 / 몰질량 × (6.022×10^23) 1 g당 분자수 = 1 / 몰질량 × (6.022×10^23) 몰질량의 단위 = g/mol 이므로, 1 / 몰질량 = 1 / (g/mol) = mol/g 아보가드로수의 단위 = 개/mol 이므로, 즉, N_A = 6.022×10^23 개/mol 이므로, ( 참고 https://ywpop.tistory.com/6475 ) 1 / 몰질량 × (6.022×10^23) = (mol/g) × (개/mol) = 개/g ---> 1 g당 분자수 예) 물 분자, H2O의 몰질량 = 18 g/mol 이므로, 1 / 18 × (6.022×10^23) = 3.345556×10^22 개/g ---> 물 분자 1 g당 분자수 그럼, 물 분자 18 g의 분자수는? (= 물 분자 1 mol의 분자수는?) 18 g × (3.3455.. 2024. 3. 26.
C2H4 4.66 g yield 89.4% C2H5Cl mass C2H4 4.66 g yield 89.4% C2H5Cl mass Ethylene reacts with hydrogen chloride to form ethyl chloride: C2H4(g) + HCl(g) → C2H5Cl(g) Calculate the mass of ethyl chloride formed if 4.66 g of ethylene reacts with an 89.4 percent yield. > C2H4의 몰질량 = 28.05 g/mol > C2H5Cl의 몰질량 = 64.51 g/mol C2H4(g) + HCl(g) → C2H5Cl(g) 4.66 g C2H4 반응 시, 생성되는 C2H5Cl의 질량을 질량비로 계산하면, ( 참고: 질량비로 계산 https://ywpop.tistory.com/.. 2024. 3. 26.
CO2 4.0% LiOH Li2CO3 7 20. L 25000 g LiOH 0.0010 g/mL CO2 4.0% LiOH Li2CO3 7 20. L 25000 g LiOH 0.0010 g/mL The space shuttle environmental system handles excess CO2 (which astronauts breathe out; it is 4.0% by mass of exhaled air) by reacting it with lithium hydroxide, LiOH, pellets to form lithium carbonate, Li2CO3, and water. If there are 7 astronauts on board the shuttle, and each exhales 20. L of air per minute, how long could clean air be gen.. 2024. 3. 26.
3.5% sulfur by weight yield H2SO4 rainfall pH 3.90 3.5% sulfur by weight yield H2SO4 rainfall pH 3.90 Acid rain was at one time an important point of contention between the United States and Canada. Much of this acid was the result of the emission of sulfur oxides by coal-fired electricity generating plants in southern Indiana and Ohio. These sulfur oxides, when dissolved in rainwater, formed sulfuric acid and hence “acid rain”. How many metric .. 2024. 3. 25.
산소 원자 2 mol의 개수 산소 원자 2 mol의 개수 1 mol = 6.02×10^23 개 이므로, ( 참고 https://ywpop.tistory.com/6475 ) 2 mol × (6.02×10^23 개 / 1 mol) = 2 × (6.02×10^23) = (2×6.02)×10^23 = 12.04×10^23 = 1.204×10^24 개 답: 1.204×10^24 개 [참고] 귤 1 box에 귤 602개가 들어있다. 귤 2 box에 들어있는 귤의 개수는? 1 box = 602개 ---> 1 box = 6.02×10^2 개 이므로, 2 box × (6.02×10^2 개 / 1 box) = 2 × (6.02×10^2) = (2×6.02)×10^2 = 12.04×10^2 = 1.204×10^3 개 [키워드] 산소 원자 아보가 기준 2024. 3. 24.
미정계수법 CrO3 → Cr2O3 + O2 미정계수법 CrO3 → Cr2O3 + O2 [참고] 미정계수법 [ https://ywpop.tistory.com/3902 ] aCrO3 → bCr2O3 + cO2 > Cr의 개수: a = 2b ... (1) > O의 개수: 3a = 3b + 2c ... (2) b = 1 이라 두면, from (1)식, a = 2 from (2)식, 6 = 3 + 2c c = 3/2 계수를 정리하면, > a = 2 > b = 1 > c = 3/2 계수는 정수이어야 하므로, 각 계수에 2씩 곱하면, > a = 4 > b = 2 > c = 3 aCrO3 → bCr2O3 + cO2 ---> 4CrO3 → 2Cr2O3 + 3O2 [키워드] CrO3 → Cr2O3 + O2 기준, 4CrO3 → 2Cr2O3 + 3O2 기준 2024. 3. 21.
미정계수법 Fe + O2 → Fe2O3 미정계수법 Fe + O2 → Fe2O3 [참고] 미정계수법 [ https://ywpop.tistory.com/3902 ] aFe + bO2 → cFe2O3 > Fe의 개수: a = 2c ... (1) > O의 개수: 2b = 3c ... (2) c = 1 이라 두면, from (1)식, a = 2 from (2)식, b = 3/2 계수를 정리하면, > a = 2 > b = 3/2 > c = 1 계수는 정수이어야 하므로, 각 계수에 2씩 곱하면, > a = 4 > b = 3 > c = 2 aFe + bO2 → cFe2O3 ---> 4Fe + 3O2 → 2Fe2O3 [ 관련 글 https://ywpop.blogspot.com/2023/11/redox-fe-o2-fe2o3.html ] redox Fe + O2.. 2024. 3. 21.
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