Question

d. How many moles of O2 must react to produce 0.250 mol of NO? e. How many grams of oxygen are needed to react with 105 g of nitrogen? 8. Hydrazine reacts with oxygen according to the unbalanced equation below: N2H4 (1) + O2(g) → NO2 (g)+ H2O(g) a. If 75.0 kg of hydrazine reacted with 75.0 kg of oxygen, which is the limiting reagent? b. How many kg of NO2 are produced from the reaction of 75.0 kg of the limiting reagent?

Answer 1

Answer:

Explanation:

A mole ratio is ​the ratio between the amounts in moles of any two compounds involved in a chemical reaction. The mole ratio can be determined by examining the coefficients in front of formulas in a balanced chemical equation.

Step 1: write the balanced chemical equation.

N₂H₄ (l) + 3 O₂ (g)  ---->2 NO₂ (g) + 2 H₂O (g)

Step 2: calculate the moles of N₂H_{4 ,} O₂

Moles of N₂H₄ = mass given / molar mass = ( 75000 g / 32.0452 g/mol ) = 2340.4 mol

Moles of O₂ = (75000 g / 32 g/mol ) = 2343.75 mol [ note: 1 kg = 1000 g ]

(a) Step 3: Determine the limiting reagent

N₂H₄ (l) + 3 O₂ (g)  ---->2 NO₂ (g) + 2 H₂O (g)

According to the reaction we need
3 mol of O₂ requires for 1 mol of N₂H₄
so, for 2343.75 mol of O₂ we will need =(1 mol of N₂H₄ / 3 mol of O₂)× 2343.75 mol of O₂ = 781.25  mol of N₂H₄
Since we need only 781.25 mol of N₂H₄ hence  N₂H₄ is in excess ( since given = 2340.4 mol )

and   O₂ is limiting reagent. so the amount of product will formed according to the limiting reagent ( i.e O₂ amount )

Step 4: Calculate the moles of NO₂ produced

N₂H₄ (l) + 3 O₂ (g)  ----> 2 NO₂ (g) + 2 H₂O (g)

According to the reaction:
3 mol of O₂ produce 2 mol of NO₂
so, 2343.75 mol of O₂ will produce=( 2 mol of NO₂ / 3 mol of O₂ )× 2343.75 mol of O₂ = 1562.5 mol of NO₂

Step 5: Calculation of NO₂ produced

we get moles of NO₂ that can be produced = 1562.5 mol

Mass of NO₂ produced =( moles × molar mass ) = ( 1562.5 mol × 46.0055 g/mol ) = 71883.6 g

Mass in kg = [ 71883.6 g × ( 1 kg / 1000 g ) ] = 71.8836 kg ≈   71.9 kg