Question

Someone can please help me with these chemistry questions?

Name: CHEM 3. (3.75) Limestone (CaCO,) is decomposed by heating to produce quicklime (CaO) and carbon dioxide. (a) Write the balanced chemical equation. (b) What is the mole relationship between limestone CaCO, to quicklime CaO. (c) Calculate how many grams of quicklime can be produced from 1.0 kg of limestone. (1000 g-1 kg: Ca 40.078 amu; O 15.999 amu). 4. (3.93) When heated, lithium reacts with dinitrogen to produce lithium nitride: 6Lis+Na 2LiNs (a) What is the theoretical yield of Li;N in grams when 12.3 g Li are heated with 33.6 g N? (Hint: which is the limiting reagent?) (b) If the actual yield of LiN is 5.89 g, what is the percent yield of the reaction ? (Li 6.941 amu; N 14.01 amu)

Answer 1

"Given: (a) CaCO_3 rightarrow^Delta CaO + CO_2 Calcium carbonato Calcium Oxide Carbon dioxide M: Molarmass MCaCO_3 = 100.0869 gm/mol McaO = 56.0774 gm/mol MCO_2 = 44.01 gm/mol (b) 1 mole of CaCO_3 rightarrow 1 mole of CaO (quicktime) (c) 1 kg of CaCO_3 = 1000 gm of CaCO_3 100 gm CaCO_3 rightarrow 1 mole 1000 gm "" rightarrow 10 mole 10 mole of CaCO_3 rightarrow 10 mole of CaO Therefore 10 mole of CaO rightarrow 10 times 56.0774 gm = 560.77 gm of quicktime(CaO) 6Li(s) + N_2 (s) rightarrow 2Li_3 N(s) MLi_3 N = 34.83 gm/mol M_N2 = 28.013 gm/mol MLi = 6.941 gm/mol 12.3 gm of Li = 1.772 mole, 33.6 gm of N_2 - 1.2 mole 6 mole of Li rightarrow 1 mole of N_2 1.77 mole of Li rightarrow x x = 0.2953 mole of N_2 (a) So 'Li' limiting reagent (b) 1 mold of N_2 rightarrow 2 mole of Li_3 N 2) 0.2953 rightarrow 2 times 0.2953 Li_3 N"