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.
Answer:
A SHOULD BE THE ANSWER I THINK
I think is true not sure but it might be
Explanation:
Let us assume that the ratio for the given reaction is 1:1.
Therefore, we will calculate the moles of 
as follows.
Moles of solution = molarity × volume (L)
= 0.0440 M × 0.014 L
= 0.000616 moles
Moles of excess EDTA = 0.000616 moles
Also, the initial moles of EDTA will be calculated as follows.
Total initial moles of EDTA = 0.0600 M × 0.025 L
= 0.0015
Therefore, moles of EDTA reacted with 
will be as follows.
= 0.0015 - 0.000616
= 0.00088 moles
Since, we have supposed a 1 : 1 ratio between and EDTA
.
So, moles of = 0.00088 moles
Now, we will calculate the molarity of as follows.
Molarity of solution = 
= 
= 0.015 M
Thus, we can conclude that the original concentration of the solution is 0.015 M.
