When a mixture of acetylene gas (C2H2) and oxygen is ignited and combusted, the reaction produces CO2 and H2O. What mass of acetylene is required to produce 150g of H2O?
Here is a stoichiometry problem I've posted as an example. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To find the mass of acetylene required to produce 150g of H2O, we need to use stoichiometry and the balanced chemical equation for the combustion of acetylene.
The balanced equation for the combustion of acetylene (C2H2) is:
2 C2H2 + 5 O2 → 4 CO2 + 2 H2O
From the equation, we can see that for every 2 moles of acetylene (C2H2) burned, 2 moles of water (H2O) is produced.
Step 1: Calculate the molar mass of water (H2O).
The molar mass of H2O = (2 × 1.008 g) + (1 × 16.00 g) = 18.016 g/mol
Step 2: Convert the mass of water (150 g) to moles.
Moles of H2O = mass of H2O / molar mass of H2O
Moles of H2O = 150 g / 18.016 g/mol ≈ 8.33 mol
Step 3: Use the stoichiometric ratio to find the moles of acetylene (C2H2) required.
From the balanced equation, we can see that 2 moles of acetylene (C2H2) produce 2 moles of water (H2O).
So, the moles of acetylene = moles of water = 8.33 mol
Step 4: Convert moles of acetylene to mass.
Molar mass of acetylene (C2H2) = (2 × 12.01 g) + (2 × 1.008 g) = 26.04 g/mol
Mass of acetylene = moles of acetylene × molar mass of acetylene
Mass of acetylene = 8.33 mol × 26.04 g/mol ≈ 216.95 g
Therefore, approximately 216.95 grams of acetylene is required to produce 150 grams of water (H2O) in this reaction.