<span>The answer is 200 mol of water.
The balanced reaction is 2(H2) + (O2) = 2(H2O)
The limiting reactant is O2 as it will be completely consumed first, before hydrogen gas. Hydrogen gas would need at least 105 mol oxygen gas to be consumed; in excess of the 100 mol O2.
Looking at the stoichiometric coefficients, the ratio between water and oxygen is 2:1.
Therefore, the water produced would be 200 moles.</span>
como se denomina el proceso utilizado para descomponer el agua
A. Protons, neutrons, and electrons
NF3– 0.94– third
NCl3–0.12– second
NBr3–0.08– first
CF4–1.43– fourth
NBr3—NCl3—NF3—CF4
Lowest. Highest
Answer:
108.6 g
Explanation:
- 2NaN₃(s) → 2Na(s) + 3N₂(g)
First we use the <em>PV=nRT formula</em> to <u>calculate the number of nitrogen moles</u>:
- R = 0.082 atm·L·mol⁻¹·K⁻¹
- T = 0 °C ⇒ 0 + 273.2 = 273.2 K
<u>Inputting the data</u>:
- 1.00 atm * 56.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.2 K
Then we <u>convert 2.5 moles of N₂ into moles of NaN₃</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:
- 2.5 mol N₂ * = 1.67 mol NaN₃
Finally we <u>convert 1.67 moles of NaN₃ into grams</u>, using its <em>molar mass</em>:
- 1.67 mol * 65 g/mol = 108.6 g