Question

Metallic silver can be obtained from sulfide ores by first roasting the sulfates (Ag_2S, ) to produce silver sulfates(Ag_2SO_4) which are leached from the ore using boiling water^11 Next the silver sulfate (Ag_2SO_4), is reacted with copper (Cu)to produce copper sulfate (CuSO_4) and silver (Ag) as illustrated in Figure 7.35. The product the second separator contains 90% (by mass) silver and 10% copper. The percent of excess in Stream #1 defined by {percent of excess copper} = (molar flow rate of copper entering in the feed stream) - (molar rate of consumption of copper in the reactor)/(molar rate of consumption of copper in the reactor) times 100 and the conversion of silver sulfate is defined by C = {molar rate of consumption of Ag_2SO_4}/{molar flow rate of Ag_2SO_4 entering the reactor} = -R_Ag_2SO_4/(M_Ag_2SO_4)_3 For the conditions given, what is the percent of excess copper? If the conversion is given by C = 0.75. what is m_6/m_5?

Answer 1

Copper h the equal acasnqchi m

2 Sehara o by maSS lot in, ed moks flow Yak φ Aasa, SO Polar ak φ production φ 5 = axo-75 mol s

62 1o 62 o lo go 62 90

'h He feed steam = 63S ニ 1.03346 mol s~! かe oacfDS o.TS Mol s-1 By uin O Porlent 03346-o-75 O 7S fe)cent- Mas S m5 433 ms 80