<span>6.03 moles.
1 molecule of butane contains 4 carbon atoms and ten hydrogen atoms.
The molar mass is 4 times the atomic mass of carbon, 12 g/mol, plus 10 times the atomic weight of hydrogen, 1 g/mol.
Molar mass = 4 * 12 g/mol + 10 * 1 g/mol = 58 g/mol.
This means that 1 mole of butane has a mass of 58 g.
To figure out how many moles are in a sample of butane, divide the mass of sample in grams by 58 grams
Number of moles in sample = 350 g / 58 g/mol = 6.03 moles.</span>
Answer:
92.04%
Explanation:
Given:
Mass of CO₂ obtained = 53.0 grams
Mass of calcium carbonate heated = 1.31 grams
Now,
the molar mass of the calcium carbonate = 100.08 grams
The number of moles heated in the problem = Mass / Molar mass
= (1.31 grams) / (100.08 grams/moles)
= 0.013088 moles
now,
1 mol of calcium carbonate yields 1 mol of CO₂
thus,
0.013088 moles of calcium carbonate will yield = 0.013088 mol of CO₂
now,
Theoretical mass of 0.013088 moles of CO₂ will be
= Number of moles × Molar mass of CO₂
= 0.013088 × 44 = 0.5758 grams
Thus, the percent yield for this reaction = 
or
the percent yield for this reaction = 
or
the percent yield for this reaction = 92.04%