Like if you imagine making something you create a hypothesis about was to make that happen
Answer:1) Volume of 
required is 55.98 mL.
2) 0.62577 grams of 
is produced.
Explanation:
1) Molarity of 
Volume of 
Molarity of 
Volume of 
According to reaction, 1 mole of 
reacts with 3 mole of , then, 0.0041985 moles of will react with:
moles of that is 0.0125955 moles.
Volume of required is 55.98 mL.
2)
Number of moles of 
According to reaction, 3 moles of gives 1 mole of , then 0.004485 moles of will give:
moles of that is 0.001495 moles.
Mass of =
Moles of × Molar Mass of
= 0.001495 moles × 418.58 g/mol = 0.62577 g
0.62577 grams of is produced.
Yeah so you have to start of with converting your first two values into moles (forget the third one)
97.5 g NO * 1 mol/30.01 g NO = 3.25 moles NO
88.0 g O2 * 1 mol/16.00 g O2 = 5.5 moles O2
now we can find the limiting reactant by checking for the amount of product each reactant should give us by using molar ratios
3.25 mol NO * 2 mol NO2/2 mol NO = 3.25 mol NO2
5.5 mol O2 * 2 mol NO2/ 1 mol O2 = 11
so NO is the limiting reactant since it produces less product/gets used up quicker
3.25 mol NO * 2 mol NO2/2molNO = 3.25 mol NO2
so this is our theoretical yield and the question provides us with the actual yield (2.68 moles). since the actual yield is given in moles, we don't have to convert to grams. our percent yield formula goes like: actual yield/theoretical yield * 100
2.68 mol/3.25 mol * 100 = 82.46%
