Question

Solving stoichiometry problems that involve mass Now that you know the molar masses of the relevant compounds, you are ready to start solving stoichiometry problems. In general, the typical strategy is Convert from grams of compound X to moles of compound X using the molar mass of compound X. Convert from moles of compound X to moles of compound Y using the coefficients in the balanced chemical equation. Convert from moles of compound Y to grams of compound Y using the molar mass of compound Y. Part B Calculate the mass of water produced when 9.56 g of butane reacts with excess oxygen.

Answer 1

14.85 g of water

Explanation:

Calculate the mass of water produced when 9.56 g of butane reacts with excess oxygen.

Here we have the full oxidation of butane (C₄H₁₀) which produce carbon dioxide and water.

C₄H₁₀ + (13/2) O₂ → 4 CO₂ + 5 H₂O

number of moles = mass / molecular weight

number of moles of butane = 9.56 / 58 = 0.165 moles

Now, taking in account the chemical reaction, we devise the following reasoning:

if         1 mole of butane produces 5 moles of water

then    0.165 moles of butane produces X moles of water

X = (0.165 × 5) / 1 = 0.825 moles of water

mass = number of moles × molecular weight

mass of water = 0.825 × 18 = 14.85 g

Learn more about:

oxidation of hydrocarbons

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