Respuesta :
Answer:
We need 6.65 L of NH3 to react with 10.0 L of NO
Explanation:
Step 1: Data given
volume of NO = 10.0 L
at STP 1 mol has a volume of 22.4 L
Step 2: The balanced equation
4NH3+6NO → 5N2+6H2O
Step 3: Calculate moles of NO
Since 22.4 L = 1 mol
10.0 L = 0.446 mol
Step 4: Calculate moles of NH3
For 6 moles NO, we need 4 moles of NH3 to produce 5 moles of N2 and 6 moles of H2O
for 0.446 moles of NO, we have 4/6 * 0.446 = 0.297 moles of NH3
Step 5: Calculate volume of NH3
1 mol = 22.4 L
0.297 mol = 22.4 * 0.297 = 6.65 L
We need 6.65 L of NH3 to react with 10.0 L of NO
Answer:
6.67 L
Explanation:
The correct equation is:
[tex]4NH_{3}+6NO[/tex] ⇒ [tex]5N_{2}+6H_{2}O[/tex]
4 moles of [tex]NH_{3}[/tex] reacts completely with 6 moles of NO to give 5 moles of [tex]N_{2}[/tex].
At standard temperature and pressure volume is directly proportional to number of moles of gas.
Hence we can say that :
4L of [tex]NH_{3}[/tex] reacts with 6L of NO.
Then to get 10L of NO [tex]\frac{4}{6}\times10[/tex] L of [tex]NH_{3}[/tex] should react.
=[tex]\frac{20}{3}=6.67[/tex]L
