Question

If 3.56g of butane is burned underneath a cup holding 1.00L of water at 21.0

Answer 1

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

Moles of butane $=\frac{3.56} {58.12}=0.06moles$

Moles of water$=\frac{10}{2}×0.06=0.3moles$

Molarity of water$=\frac{0.3}{1.00}=0.3M$