Jawaban:
massa Fe = 1,68 gr
volume H2 = 0,448 L
Penjelasan:
setarakan reaksinya terlebih dahulu
2Fe+3H2SO4→Fe2(SO4)3+3H2
hitung mol H2SO4
n H2SO4 = M.V
= 0,2x100
= 20 mmol=0,02mol
n Fe = x 0,02 mol = 0,03 mol
n Fe =
massa Fe = 0,03x 56 = 1,68 gr
n H2 = n H2SO4 = 0,02 mol
V H2 = n H2x22,4 (karena STP)
= 0,02x22,4
= 0,448 L
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Jawaban:
massa Fe = 1,68 gr
volume H2 = 0,448 L
Penjelasan:
setarakan reaksinya terlebih dahulu
2Fe+3H2SO4→Fe2(SO4)3+3H2
hitung mol H2SO4
n H2SO4 = M.V
= 0,2x100
= 20 mmol=0,02mol
n Fe = x 0,02 mol = 0,03 mol
n Fe =
massa Fe = 0,03x 56 = 1,68 gr
n H2 = n H2SO4 = 0,02 mol
V H2 = n H2x22,4 (karena STP)
= 0,02x22,4
= 0,448 L