Step one calculate the moles of each element
that is moles= % composition/molar mass
molar mass of Ca = 40g/mol, S= 32 g/mol , O= 16 g/mol
moles of Ca = 29.4 /40g/mol=0.735 moles, S= 23.5/32 =0.734 moles, O= 47.1/16= 2.94 moles
calculate the mole ratio by dividing each mole with smallest mole that is 0.734
Ca= 0.735/0.734= 1, S= 0.734/0.734 =1, O = 2.94/ 0.734= 4
therefore the emipical formula = CaSO4
Answer:
Actual yield = 86.5g
Explanation:
Percent yield = 82.38%
Theoretical yield = 105g
Actual yield = x
Equation of reaction,
CaCO₃ + HCl → CaCl₂ + CO₂ + H₂O
Percentage yield = (actual yield / theoretical yield) * 100
82.38% = actual yield / theoretical yield
82.38 / 100 = x / 105
Cross multiply and make x the subject of formula
X = (105 * 82.38) / 100
X = 86.499g
X = 86.5g
Actual yield of CaCl₂ is 86.5g
