N₂+3H₂⇒ 2NH₃
m(NH₃)=1250+225*2=1700 grams
N₂ is the limiting <span>reagent.
1250 grams are</span><span> left when the maximum amount of ammonia is formed.</span>
The usual units of density are g/cm.
You may have also seen g/mL used for density. Keep in mind that 1 cm = 1 mL.
Answer:
The answer is
<h3>3.75 × 10²⁴ atoms of Al</h3>
Explanation:
To find the number of atoms of Al given it's number of moles we use the formula
<h3>N = n × L</h3>
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question
n = 6.23 mol
We have
N = 6.23 × 6.02 × 10²³
We have the final answer as
<h3>3.75 × 10²⁴ atoms of Al</h3>
Hope this helps you
