Answer:
46 g
Explanation:
The balanced equation of the reaction between O and NO is
2 NO + O₂ ⇔ 2 NO₂
Now, you need to find the limiting reagent. Find the moles of each reactant and divide the moles by the coefficient in the equation.
NO: (80 g)/(30.006 g/mol) = 2.666 mol
(2.666 mol)/2 = 1.333
O₂: (16 g)/(31.998 g/mol) = 0.500 mol
(0.500 mol)/1 = 0.500 mol
Since O₂ is smaller, this is the limiting reagent.
The amount of NO₂ produced will depend on the limiting reagent. You need to look at the equation to determine the ratio. For every mole of O₂ reacted, 2 moles of NO₂ are produced.
To find grams of NO₂ produced, multiply moles of O₂ by the ratio of NO₂ to O₂. Then, convert moles of NO₂ to find grams.
0.500 mol O₂ × (2 mol NO₂/1 mol O₂) = 1.000 mol NO₂
1.000 mol × 46.005 g/mol = 46.005 g
You will produce 46 g of NO₂.
The balanced equation for the above reaction is as follows
C₆H₁₂O₆(s) + 6O₂(g) --> 6H₂O(g) + 6CO₂<span>(g)
the limiting reactant in the equation is glucose as the whole amount of glucose is used up in the reaction.
the amount of </span>C₆H₁₂O₆ used up - 13.2 g
the number of moles reacted - 13.2 g/ 180 g/mol = 0.073 mol
stoichiometry of glucose to CO₂ - 1:6
then number of CO₂ moles are - 0.073 mol x 6 = 0.44 mol
As mentioned this reaction takes place at standard temperature and pressure conditions,
At STP 1 mol of any gas occupies 22.4 L
Therefore 0.44 mol of CO₂ occupies 22.4 L/mol x 0.44 mol = 9.8 rounded off - 10.0 L
Answer is B) 10.0 L CO₂
This might help you hope so
It would be considered an element as nothing has been added or taken away from it
The purpose of the uninoculated control tubes used in this test is that two uninoculated control tubes are needed to show the results of the medium in both aerobic and anaerobic environments. It is used to show it is sterile and also as a color comparison, used also to show that the medium remains green under both conditions.
