Answer:
5.9 kg
Explanation:
We must work backwards from the second step to work out the mass of oxygen.
1. Second step
Mᵣ: 55.84
Fe₂O₃ + 3CO ⟶ 2Fe + 3CO₂
m/kg: 7.0
(a) Moles of Fe

(b) Moles of CO

However, this is the theoretical yield.
The actual yield is 72. %.
We need more CO and Fe₂O₃ to get the theoretical yield of Fe.
(c) Percent yield

We must use 261 mol of CO to get 7.0 kg of Fe.
2. First step
Mᵣ: 32.00
2C + O₂ ⟶ 2CO
n/mol: 261
(a) Moles of O₂

(b) Mass of O₂

However, this is the theoretical yield.
The actual yield is 71. %.
We need more C and O₂ to get the theoretical yield of CO.
(c) Percent yield

We need 5.9 kg of O₂ to produce 7.0 kg of Fe.
Answer:
Gold: 1.1 x 10²² atoms/cm³
Silver: 4.8 x 10²² atoms/cm³
Explanation:
100 g of the alloy will have 29 g of Au and 71 g of Ag.
19.32 g Au ____ 1 cm³
29 g Au ______ x
x = 1.5 cm³
10.49 g Ag ____ 1 cm³
71 g Ag _______ y
y = 6.8 cm³
The total volume of 100g of the alloy is x+y = 8.3 cm³.
Gold:
196.97 g Au____ 6.022 x 10²³ atoms Au
29 g Au _______ w
w = 8.9 x 10²² atoms Au
8.9 x 10²² atoms Au ____ 8.3 cm³
A ____ 1 cm³
A = 1.1 x 10²² atoms Au
Silver:
107.87 g Ag____ 6.022 x 10²³ atoms Ag
71 g Ag _______ w
w = 4.0 x 10²³ atoms Ag
4.0 x 10²³ atoms Ag ____ 8.3 cm³
B ____ 1 cm³
B = 4.8 x 10²² atoms Ag