The reaction equation is:
<span>2CuO(s) + C(s) </span>→ <span>2Cu(s) + CO</span>₂<span>(g)
First, we determine the number of grams present in one ton of copper oxide. This is:
1 ton = 9.09 x 10</span>⁵ g
We convert this into moles by dividing by the molecular mass of copper oxide, which is:
9.09 x 10⁵ / 79.5 = 11,434 moles
Each mole of carbon reduces two moles of copper oxide, so the moles of carbon required are:
11,434 / 2 = 5,717 moles of Carbon required
The mass of carbon is then:
5,717 x 12 = 68,604 grams
The mass of coke is:
68,604 / 0.95 = 72,214 g
The mass of coke required is 7.22 x 10⁴ grams
Answer:
The equilibrium concentration of NO is 0.001335 M
Explanation:
Step 1: Data given
The equilibrium constant Kc is 0.0025 at 2127 °C
An equilibrium mixture contains 0.023M N2 and 0.031 M O2,
Step 2: The balanced equation
N2(g) + O2(g) ↔ 2NO(g)
Step 3: Concentration at the equilibrium
[N2] = 0.023 M
[O2] = 0.031 M
Kc = 0.0025 = [NO]² / [N2][O2]
Kc = 0.0025 = [NO]² / (0.023)(0.031)
[NO] = 0.001335 M
The equilibrium concentration of NO is 0.001335 M
1 mole of N2 produces 2 moles of NH3
OR...
14 x 2 grams of N2 produces 2(14 +3) grams of NH3
1 gram of N2 produces 34/28 grams of NH3
therefore, 56 grams produce (34/28 )x 56 =68 grams of NH3
the answer thus would be 68 grams of NH3
Answer:
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