What mass of hydrogen bromide is required to interact with ethylene with a volume of 44.8 liters?

univerkov | December 19, 2020 | education

С2Н4 + НBr -> C2H5Br Let us find n С2Н4: n = V / Vm; Vm = 22.4 l / mol n (C2H4) = 44.8 l / 22.4 l / mol = 2 mol Find n HBr n (HBr) = n (C2H4) = 2 mol Find m HBr m = M * nm = 2 mol * 81 g / mol = 162 g
Answer: 162 g

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