Answer:
1.57 mol NaN₃
Explanation:
- 2 NaN₃ (s) → 2 Na (s) + 3 N₂ (g)
First we <u>use PV=nRT to calculate the number of N₂ moles that need to be produced</u>:
- R = 0.082 atm·L·mol⁻¹·K⁻¹
- T = 23.7 °C ⇒ 23.7 + 273.16 = 296.86 K
<u>Inputing the data</u>:
- 1.07 atm * 53.4 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 296.86
And <u>solving for n</u>:
Finally we <u>convert N₂ moles into NaN₃ moles</u>, using <em>the stoichiometric coefficients of the balanced reaction</em>:
- 2.35 mol N₂ *
= 1.57 mol NaN₃
Well, we need to find the ratio of Al to the other reactant.
Al:HCl = 1:3
--> this means that for every 1 Al used, you have to use 3 HCl.
6*3 = 18 moles of HCl needed to fully react with 6 moles of Al. Since 13<18, HCL is the limiting reactant.
The ratio of HCl:AlCl = 3:1
13/3 = 4.3333...
The final answer is HCl is the limiting reactant with 4.3 moles of AlCl3 able to be produced.
Hope this helps!!! :)
Answer:
When the concentration of all the reactants increases, more molecules or ions interact to form new compounds, and the rate of reaction increases.
Explanation: