Carbon dioxide is removed from the atmosphereof space capsules by reaction with
ID: 749856 • Letter: C
Question
Carbon dioxide is removed from the atmosphereof space capsules by reaction with a solid metal hydroxide. Theproducts are water and the metal carbonate. (a) Calculate the mass of CO2 thatcan be removed by reaction with 3.12kg of lithium hydroxide. g (b) How many grams of CO2 can be removed by 1.14 g of each of the following substances:lithium hydroxide, magnesium hydroxide, and aluminum hydroxide? LiOH g Mg(OH)2 g Al(OH)3 g Thanks you so so much!Explanation / Answer
You need to set up balanced equations for each of the reactions involving carbon dioxide: CO2 + 2LiOH --> H2O + Li2CO3 CO2 + Mg(OH)2 --> H2O + MgCO3 3CO2 + 2Al(OH)3 --> 3H2O + Al2(CO3)2 (a) So, let's figure out the number of moles of 3.88 kg LiOH: moles = (3.88 kg x 1000 g/kg) / (23.95 g/mole LiOH) = 162.00 moles. Since 2 moles of LiOH react with one mole of CO2, moles CO2 = 1/2 x 162.00 = 81.00 moles Therefore, grams CO2 absorbed = 81.00 moles x 44 g/mole CO2 = 3564 gm = 3560 grams (three sig figs). (b) The same procedure can be used for the others, too. Find the moles of each hydroxide compound, convert that to moles of CO2 via the above equations, and turn those into grams CO2: LiOH: [(3.12 g LiOH)/ (23.95 g/mole LiOH)] x [(1 mole CO2) / (2 moles LiOH)] x 44 g/mole CO2 = 2.87 g CO2 Mg(OH)2: [(3.12 g Mg(OH)2)/ (58.33 g/mole Mg(OH)2)] x [(1 mole CO2) / (1 moles Mg(OH)2)] x 44 g/mole CO2 = 2.35 g CO2 Al(OH)3: [(3.12 g Al(OH)3)/ (78.00 g/mole Al(OH)3)] x [(3 mole CO2) / (2 moles Al(OH)3)] x 44 g/mole CO2 = 2.64 g CO2