Final answer:
To find the mass of NO produced from 30.0 g of NO2, we convert the mass of NO2 to moles, use the mole ratio to find moles of NO, and finally convert these moles to mass, resulting in 6.51 g NO.
Step-by-step explanation:
Stoichiometry of Nitrogen Dioxide and Water Reaction
The problem involves calculating the mass of nitric oxide (NO) that will be produced when 30.0 g of nitrogen dioxide (NO2) reacts with excess water. This is a stoichiometry problem where we'll use the chemical equation 3NO2(g) + H2O(l) → 2HNO3(g) + NO(g) as a reference to find the molar relationship between NO2 and NO. To solve this, we will first convert the mass of NO2 into moles using its molar mass (46.01 g/mol), then apply the mole ratio from the balanced equation, and finally convert the moles of NO back into grams using NO's molar mass (30.01 g/mol).
Step 1: Determine moles of NO2.
Moles of NO2 = mass of NO2 / molar mass of NO2
= 30.0 g / 46.01 g/mol = 0.652 mol NO2
Step 2: Use the mole ratio to determine moles of NO.
Moles of NO = moles of NO2 × (1 mole NO / 3 moles NO2)
= 0.652 mol × (1/3) = 0.217 mol NO
Step 3: Convert moles of NO to mass.
Mass of NO = moles of NO × molar mass of NO
= 0.217 mol × 30.01 g/mol = 6.51 g NO