4Fe+3O2 I really have no idea how to do this.. HELP
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Answer:
X 86 206
Explanation:
Radioactive atoms are nuclei that can under go disintegration to emit either an alpha particle, beta particle or gamma radiation. The process could be spontaneous or stimulated.
When a radioactive atom R 88 210 emits alpha particle, it would produce an element with atomic number 86 and mass number 206 i.e X 86 206. An alpha particle is usually a helium nucleus.
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+
+ energy
Percentage yield = (actual yield / theoretical yield) x 100%
The balanced equation for the decomposition is,
2Na₃(CO₃)(HCO₃)·2H₂O(s) → 3Na₂CO₃(s) + CO₂(g) + 5H₂<span>O(g)
The stoichiometric ratio between </span>Na₃(CO₃)(HCO₃)·2H₂O(s) and Na₂CO₃(s) is 2 : 3
The decomposed mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 1000 kg
= 1000 x 10³ g
Molar mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 226 g mol⁻¹
moles of Na₃(CO₃)(HCO₃)·2H₂O(s) = mass / molar mass
= 1000 x 10³ g / 226 g mol⁻¹
= 4424.78 mol
Hence, moles of Na₂CO₃ formed = 4424.78 mol x
= 6637.17 mol
Molar mass of Na₂CO₃ = 106 g mol⁻¹
Hence, mass of Na₂CO₃ = 6637.17 mol x 106 g mol⁻¹
= 703540.02 g
= 703.540 kg
Hence, the theoretical yield of Na₂CO₃ = 703.540 kg
Actual yield of Na₂CO₃ = 650 kg
Percentage yield = (650 kg / 703.540 kg) x 100%
= 92.34%
The balanced equation for the decomposition is,
2Na₃(CO₃)(HCO₃)·2H₂O(s) → 3Na₂CO₃(s) + CO₂(g) + 5H₂<span>O(g)
The stoichiometric ratio between </span>Na₃(CO₃)(HCO₃)·2H₂O(s) and Na₂CO₃(s) is 2 : 3
The decomposed mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 1000 kg
= 1000 x 10³ g
Molar mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 226 g mol⁻¹
moles of Na₃(CO₃)(HCO₃)·2H₂O(s) = mass / molar mass
= 1000 x 10³ g / 226 g mol⁻¹
= 4424.78 mol
Hence, moles of Na₂CO₃ formed = 4424.78 mol x
= 6637.17 mol
Molar mass of Na₂CO₃ = 106 g mol⁻¹
Hence, mass of Na₂CO₃ = 6637.17 mol x 106 g mol⁻¹
= 703540.02 g
= 703.540 kg
Hence, the theoretical yield of Na₂CO₃ = 703.540 kg
Actual yield of Na₂CO₃ = 650 kg
Percentage yield = (650 kg / 703.540 kg) x 100%
= 92.34%