I think the answer is D (number 4)
47% yield.
First, let's determine how many moles of ethane was used and how many moles of CO2 produced. Start with the respective atomic weights.
Atomic weight carbon = 12.0107
Atomic weight hydrogen = 1.00794
Atomic weight oxygen = 15.999
Molar mass C2H6 = 2 * 12.0107 + 6 * 1.00794 = 30.06904 g/mol
Molar mass CO2 = 12.0107 + 2 * 15.999 = 44.0087 g/mol
Moles C2H6 = 8 g / 30.06904 g/mol = 0.266054387 mol
Moles CO2 = 11 g / 44.0087 g/mol = 0.249950578 mol
Looking at the balanced equation, for every 2 moles of C2H6 consumed, 4 moles of CO2 should be produced. So at 100% yield, we should have 0.266054387 / 2 * 4 = 0.532108774 moles of CO2. But we only have 0.249950578 moles, or 0.249950578 / 0.532108774 = 0.46973587 =
46.973587% of what was expected.
Rounding to 2 significant figures gives 47% yield.
Answer:
74.9%.
Explanation:
Relative atomic mass data from a modern periodic table:
- Ca: 40.078;
- C: 12.011;
- O: 15.999.
What's the <em>theoretical</em> yield of this reaction?
In other words, what's the mass of the CO₂ that should come out of heating 40.1 grams of CaCO₃?
Molar mass of CaCO₃:
.
Number of moles of CaCO₃ available:
.
Look at the chemical equation. The coefficient in front of both CaCO₃ and CO₂ is one. Decomposing every mole of CaCO₃ should produce one mole of CO₂.
.
Molar mass of CO₂:
.
Mass of the 0.400655 moles of 
expected for the 40.1 grams of CaCO₃:
.
What's the <em>percentage</em> yield of this reaction?
.
Answer:
Block Y is heavier than Block X.
Explanation:
