Answer:
The percent yield of this reaction is 70%
Explanation:
The reaction is: N₂ + 3H₂ → 2NH₃
We only have the mass of H₂, so we assume that N₂ is in excess
We convert the mass to moles, to work with the reaction:
450 g . 1mol / 2 g = 225 moles
Ratio is 2:3. 3 moles of H₂ can produce 2 moles of ammonia
Therefore 225 moles of H₂ will produce (225 .2)/ 3 = 150 moles
This is the 100% yield reaction → We convert the moles of NH₃ to mass
150 mol . 17g /1mol = 2550 g
Percent yield = (Produced yield/Theoretical yield) .100
Percent yield = (1575g/2550g) . 100 = 70%
Answer:
You will get 5.0 g of hydrogen.
Explanation:
As with any stoichiometry problem, we start with the balanced equation.
Sn
l
+
2HF
→
SnF
2
+
H
2
Moles of H
2
=
2.5
mol Sn
×
1 mol H
2
1
mol Sn
=
2.5 mol H
2
Mass of H
2
=
2.5
mol H
2
×
2.016 g H
2
1
mol H
2
=
5.0 g H
2
Answer:
To prepare 50L of 32% solution you need: 11L of 30% solution, 22L of 50% solution and 17L of 10% solution.
Explanation:
A 32% solution of acid means 32L of acid per 100L of solution. As the chemist wants to make a solution using twice as much of the 50% solution as of the 30% solution it is possible to write:
2x*50% + x*30% + y*10% = 50L*32%
<em>130x + 10y = 1600 </em><em>(1)</em>
<em>-Where x are volume of 30% solution, 2x volume of 50% solution and y volume of 10% solution-</em>
Also, it is possible to write a formula using the total volume (50L), thus:
<em>2x + x +y = 50L</em>
<em>3x + y = 50L </em><em>(2)</em>
If you replace (2) in (1):
130x + 10(50-3x) = 1600
100x + 500 = 1600
100x = 1100
<em>x = 11L -Volume of 30% solution-</em>
2x = 22L -Volume of 50% solution-
50L - 22L - 11L = 17 L -Volume of 10% solution-
I hope it helps!
