The interaction of bromine weighing 160 g with hydrogen forms 139.32 g of hydrogen bromide.
| December 3, 2020 | education
The interaction of bromine weighing 160 g with hydrogen forms 139.32 g of hydrogen bromide. Determine the yield of this chemical reaction.
H2 + Br2 = 2HBr
n (Br2) = m \ M = 160 \ 160 = 1 mol
for ur-th p-i n (HBr) = 1 * 2 = 2 mol
m theor (HBr) = nM = 2 * 81 = 162 g
output = m (practical) \ m (theory) = 139.32 \ 162 = 0.86 (86%)
Answer: 86%
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