If 12,0 g is allowed to react with each of the 3 reactants, I know that Na2CO3 is the limiting agent. Then, if 0.226 moles of 2NaCN is produced from this equation, how many grams of C and N left unreacted??? Please give ur way... I am totally blur with this. Thank you... (answer: c=6,58 g and N2 = 8.85g)
2007-05-18 15:50:34 · 3 answers · asked by NG Yut Sei 1 in Science & Mathematics ➔ Chemistry
moles Na2CO3 = 12 / 106 = 0.113
moles C = 1
moles N2 = 0.857
we get 0.226 moles NaCN and 0.339 moles CO
Total moles C = 1 + 0.113 = 1.113
Total moles C reacted = 0.226 + 0.339 = 0.565
moles C unreacted = 1.113 - 0.565 = 0.548
g C unreacted = 12 g / mol x 0.548 = 6.58
moles N unreacted = 0.857 - 0.226 = 0.631
0.631 x 14 g/mol = 8.83 g
2007-05-18 16:04:27 · answer #1 · answered by Anonymous · 1⤊ 0⤋
moles Na2CO3 = 12 / 106 = 0.113
moles C = 1
moles N2 = 0.857
we get 0.226 moles NaCN and 0.339 moles CO
Total moles C = 1 + 0.113 = 1.113
Total moles C reacted = 0.226 + 0.339 = 0.565
moles C unreacted = 1.113 - 0.565 = 0.548
g C unreacted = 12 g / mol x 0.548 = 6.58
2007-05-18 16:10:12 · answer #2 · answered by madskillz1358 2 · 0⤊ 1⤋
c/(4*12.01 = 12/(2*22.99 + 12.01 + 3*16)
c/48.01 = 12/105.99
c = 5.436
2 - c = 6.564 g C
n/(2*14.01) = 12/105.99
n = 3.172
12 - n = 8.828 g N2
2007-05-18 16:18:15 · answer #3 · answered by Helmut 7 · 0⤊ 0⤋