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In a bromine-producing plant, how many liters of gaseous elemental bromine at 300°C and 0.855 atm are formed by the reaction of 275 g of sodium bromide and 175.6 g of sodium bromate in aqueous acid solution? (Assume that no Br₂ dissolves.)
5NaBr(aq) + NaBrO₃(aq) + 3H₂SO₄(aq) →
3Br₂(g) + 3Na₂SO₄(aq) + 3H₂O(g)

Respuesta :

5NaBr(aq) + NaBrO3(aq) + 3H2 SO4(aq)

---->3Br{2(g)}+3Na{2}SO{4(aq)}+3H{2}O_{(g)}   is the balanced equation

Given data

T=300°C=573,15K                   p=0,855atm=86632,88Pa

m(NaBr)=275g                          Mr(NaBr)=102,89g/mol

n(NaBr)=m/Mr=2,67mol             m(NaBrO3 )=175,6g

Mr(NaBrO3  )=150,89g/mol

n(NaBrO3  )=m/Mr=1,16mol

If NaBrO3  was limiting reactant we would need 5,8 mol of NaBr and we do not have that so NaBr is the limiting reactant.

n(Br2 )=   3/5*n(NaBr)=  1.6mol

pV=nR*T                                      V=n∗R∗T/p

V=1.6 ∗8.314∗ 573.15  /  86632,88

V=0.0882m3

V=88.21L

To know more about  sodium bromide and sodium bromate in aqueous solution here

brainly.com/question/12884734

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