Isopropyl alcohol (c3h7OH) burns in air according to this equation: 2C3H7OH + 9O2 -> 6CO2 + 8H2O
a) calculate the moles of oxygen needed to react with 3.40 mol C3H7OH
b) find the moles of each product formed when 3.40 mol C3H7OH reacts with oxygen

a) 15.3 mol O₂

b) 10.2 mol CO₂; 13.6 mol H₂O

Explanation:

Step 1: Write the balanced equation

2 C₃H₇OH + 9 O₂ ⇒ 6 CO₂ + 8 H₂O

Step 2: Calculate the moles of O₂ that react with 3.40 moles of C₃H₇OH

The molar ratio of C₃H₇OH to O₂ is 2:9. The moles of O₂ required are 9/2 × 3.40 mol = 15.3 mol

Step 3: Calculate the moles of CO₂ produced from 3.40 moles of C₃H₇OH

The molar ratio of C₃H₇OH to CO₂ is 2:6. The moles of CO₂ produced are 6/2 × 3.40 mol = 10.2 mol

Step 4: Calculate the moles of H₂O produced from 3.40 moles of C₃H₇OH

The molar ratio of C₃H₇OH to H₂O is 2:8. The moles of H₂O produced are 8/2 × 3.40 mol = 13.6 mol

the maximum mass of nh3 is nh3 in grams