A cellphone battery is a electrical energy
and a piston in an engine is an thermal energy
Answer : 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.
Solution : Given,
Mass of 
= 100 g
Molar mass of = 27 g/mole
Molar mass of 
= 28 g/mole
First we have to calculate moles of .
The given balanced chemical reaction is,
From the given reaction, we conclude that
2 moles of produced from 1 mole of
3.7 moles of produced from 
of
Now we have to calculate the mass of .
Mass of = Moles of × Molar mass of
Mass of = 1.85 mole × 28 g/mole = 51.8 g
Therefore, 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.
Answer:
Percent yield: 78.2%
Explanation:
Based on the reaction:
4Al + 3O₂ → 2Al₂O₃
<em>4 moles of Al produce 2 moles of Al₂O₃</em>
<em />
To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:
(Actual yield (6.8g) / Theoretical yield) × 100
Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:
4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.
As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:
0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = <em>0.0852 moles of Al₂O₃</em>,
In grams (Molar mass Al₂O₃ = 101.96g/mol):
0.0852 moles of Al₂O₃ × (101.96g / mol) =
<h3>8.7g of Al₂O₃ can be produced (Theoretical yield)</h3>
Thus, Percent yield is:
(6.8g / 8.7g) × 100 =
<h3>
78.2% </h3>
Molar mass :
Li₂S = <span>45.947 g/mol
AlCl</span>₃ = <span>133.34 g/mol
</span><span>3 Li</span>₂<span>S + 2 AlCl</span>₃<span> = 6 LiCl + Al</span>₂S₃
3 * 45.947 g Li₂S ----------> 2 * <span>133.34 g AlCl</span>₃
1.084 g Li₂S ----------------> ?
Mass Li₂S = 1.084 * 2 * 133.34 / 3 * 45.947
Mass Li₂S = 289.08112 / 137.841
Mass Li₂S = 2.0972 g
hope this helps!
