If there is a mass percentage of 24.5% gold in an ore vein. How many moles of gold can be produced from 1.00 x 103 kg of ore? Assume 100% recovery
Proportion:
1000 kg – 100 %
x kg – 24.5 %
x=1000×24.5100=245 kg=245×103 gx = \frac{1000 \times 24.5}{100} = 245 \; kg = 245 \times 10^3 \; gx=1001000×24.5=245kg=245×103g
MM(Au) = 196.967 g/mol
n(Au)=245×103196.967=1.243×103 mol=1243 moln(Au) = \frac{245 \times 10^3}{196.967} = 1.243 \times 10^3 \;mol = 1243 \;moln(Au)=196.967245×103=1.243×103mol=1243mol
Answer: 1243 mol
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