Answer:
O2 is limiting reactant
Explanation:
To find the limiting reactant we need to convert the mass of each reactant to the moles using the formula weight. And, as 1 mole of C6H12O6 reacts with 6 moles of O2, we can know wich reactant will be over first (Limiting reactant) as follows:
<em>Moles C6H12O6:</em>
650g * (1mol/180.16g) = 3.608 moles C6H12O6
<em>Moles O2:</em>
650g * (1mol/32g) = 20.31 moles O2
Now, for a complete reaction of 3.608 moles of C6H12O6 are required:
3.608 moles C6H12O6 * (6mol O2 / 1mol C6H12O6) = 21.65 moles O2
As there are just 20.31 moles of O2,
<h3>O2 is limiting reactant</h3>
Answer:
Sorry
Explanation:
Sorry this is not chemistry but I always try to answer but this time I can't I am so so sorry
The theoretical proportion is given by the balanced chemical equation:
2 mol NBr / 3 mol Na OH
Then x mol NaOH / 40 mol NBr3 = 3mol NaOH/2 mol NBr3
Solve for x, x = 40 * 3/2 = 60 mol NaOH.
Given that there are 48 mol NaOH (less than 60) this is the limitant reactant and the other is the excess reactant.
Answer: NBr3..
