Answer:
Ya'll should really practice more because things like these are easy
Explanation:
Answer:
13 mol NO
Explanation:
Step 1: Write the balanced equation
4 NH₃(g) + 5 O₂(g) ⇒ 4 NO(g) + 6 H₂O(g)
Step 2: Establish the appropriate molar ratio
According to the balanced equation, the molar ratio of O₂ to NO is 5:4.
Step 3: Calculate the number of moles of O₂ needed to produce 16 moles of NO
We will use the previously established molar ratio.
16 mol O₂ × 4 mol NO/5 mol O₂ = 13 mol NO
Answer:
the answer is longgitudinal
Answer:
2 moles
Explanation:
Hi there !!
Given volume = 44.8 litres
Molar volume = 22.4 litres
No: of moles = Given volume / Molar volume
= 44.8 / 22.4 = 2 moles
Almost all hydrocarbon 'burn' reactions involve oxygen; it's by far the most reactive substance in air.
<span>Hydrocarbon combustions always involve </span>
<span>[some hydrocarbon] + oxygen --> carbon dioxide + steam. </span>
C6H6(l) + O2 (g)--> CO2 (g)+ H2O (g)
<span>Balance carbon, six on each side: </span>
C6H6(l) + O2 (g)--> 6CO2 (g)+ H2O (g)
<span>Balance hydrogen, six on each side: </span>
C6H6(l) + O2 (g)--> 6CO2(g) + 3H2O (g)
<span>Now, we have fifteen oxygens on the right and O2 on the left. </span>
<span>Two ways to deal with that. We can use a fraction: </span>
C6H6 (l)+ (15/2)O2 (g)--> 6CO2 (g)+ 3H2O (g)
<span>Or, if you prefer to have whole number coefficients, double everything </span>
<span>to get rid of the fraction: </span>
2C6H6 (l)+ 15O2 (g)--> 12CO2 (g)+ 6H2O (g)
<span>With the SATP states thrown in... </span>
C6H6(l) + (15/2)O2(g) --> 6CO2(g) + 3H2O(g)
