40.1g of nitrogen gas is produced.
The equation given is
2 NH₃ + 3 CuO →3 Cu + N₂ + 3 H₂O
This equation is already balanced.
When 3 moles of CuO are consumed, 1 mole of nitrogen gas is produced.
We get 1 mole of nitrogen from 3 moles of copper oxide.
We need to find the number of moles of nitrogen gas produced when 4.3 moles of copper oxide are consumed.
4.3/3 x 1 = 1.433 mols
- 1.433 mols of nitrogen gas are produced
- The molar mass of nitrogen gas is 14+14 = 28g
- The amount of nitrogen gas produced in grams is 28x1.433 = 40.1g
40.1g of nitrogen gas can be made when 4.3 moles of CuO are consumed.
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Mg(OH)₂ ⇄ Mg²⁺ + 2 OH⁻
Ksp = [Mg²⁺] [OH⁻]²
6.0 x 10⁻¹⁰ = 0.10 x [OH⁻]²
[OH⁻] = 7.746 x 10⁻⁵ M
when Mg(OH)₂ 1st precipitates, [OH⁻] = 7.746 * 10⁻⁵ M
Fe(OH)₂ <—> Fe²⁺ + 2OH⁻
Ksp = [Fe²⁺] [OH⁻]²
7.9 x 10⁻¹⁶ = [Fe²⁺] x (7.746 x 10⁻⁵)²
[Fe²⁺] = 1.32 x 10⁻⁷ M
Answer: 1.32 x 10⁻⁷ M
1.2204504e-18 i think hope this helped
