Answer:
204g of NH3
Explanation:
The balanced equation for the reaction is given below:
N2 + 3H2 —> 2NH3
Next, we shall determine the number of mole NH3 produced by reacting 6moles of N2. This is illustrated below:
From the balanced equation above,
1 mole of N2 reacted to produce 2 moles of NH3.
Therefore, 6 moles of N2 will react to produce = 6 x 2 = 12 moles of NH3.
Finally, we shall convert 12 moles of NH3 to grams. This is illustrated below:
Number of mole of NH3 = 12 moles.
Molar mass of NH3 = 14 + (3x1) = 17g/mol
Mass of NH3 =..?
Mass = mole x molar mass
Mass of NH3 = 12 x 17
Mass of NH3 = 204g.
Therefore, 204g of NH3 will be produced from the reaction.
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Answer is: <span>yield of a reaction is 56,4%.
</span>Chemical reaction: PCl₃ + 3H₂O → 3HCl + H₃PO₃.
m(PCl₃) = 200 g.
m(HCl) = 91,0 g.
n(PCl₃) = m(PCl₃) ÷ M(PCl₃).
n(PCl₃) = 200 g ÷ 137,33 g/mol.
n(PCl₃) = 1,46 mol.
n(HCl) = m(HCl) ÷ M(HCl).
n(HCl) = 91 g ÷ 36,45 g/mol.
n(HCl) = 2,47 mol.
From reaction: n(PCl₃) : n(HCl) = 1 : 3.
n(HCl) = 1,46 mol · 3 = 4,38 mol.
Yield of reaction: 2,47 mol ÷ 4,38 mol · 100% = 56,4%.
The reaction produces 2.93 g H₂.
M_r: 133.34 2.016
2Al + 6HCl → 2AlCl₃ + 3H₂
<em>Moles of AlCl₃</em> = 129 g AlCl₃ × (1 mol AlCl₃/133.34 g AlCl₃) = 0.9675 mol AlCl₃
<em>Moles of H₂</em> = 0.9675 mol AlCl₃ × (3 mol H₂/2 mol AlCl₃) = 1.451 mol H₂
<em>Mass of H₂</em> = 1.451 mol H₂ × (2.016 g H₂/1 mol H₂) = 2.93 g H₂
Divide 1.25/3.93 and then multiply by 100 to get the percent. This equals 31.81%
