Final answer:
The complete reaction of 0.234 mole of O₂ will form 0.468 mole of NO₂. The complete reaction of 3.04 g of NO will form 4.652 g of NO₂.
Step-by-step explanation:
(a) To determine how many moles of NO₂ are formed by the complete reaction of 0.234 mole of O₂, we can use the balanced equation:
2 NO(g) + O₂(g) → 2 NO₂(g)
From the equation, we can see that for every 1 mole of O₂, 2 moles of NO₂ are formed. Therefore, with 0.234 mole of O₂, we can use the ratio:
0.234 mole O₂ × (2 mole NO₂/1 mole O₂) = 0.468 mole NO₂
Therefore, 0.234 mole of O₂ will form 0.468 mole of NO₂.
(b) To determine how many grams of NO₂ are formed by the complete reaction of 3.04 g of NO, we can use the molar mass of NO (30.01 g/mol) and the balanced equation:
2 NO(g) + O₂(g) → 2 NO₂(g)
First, we need to convert 3.04 g of NO to moles:
3.04 g NO × (1 mol NO/30.01 g NO) = 0.101 mol NO
From the equation, we can see that for every 2 moles of NO, 2 moles of NO₂ are formed. Therefore, with 0.101 mole of NO, we can use the ratio:
0.101 mole NO × (2 mole NO₂/2 mole NO) = 0.101 mole NO₂
Finally, we can convert the moles of NO₂ to grams using the molar mass of NO₂ (46.01 g/mol):
0.101 mol NO₂ × (46.01 g NO₂/1 mol NO₂) = 4.652 g NO₂
Therefore, 3.04 g of NO will form 4.652 g of NO₂.