The number of grams of Ba₃(PO₄)₂ that is produced is : 0.4328 gram
Given data
mass of Na₃PO₄.12H₂O = 0.629 grams
mass of BaCl₂.2H₂O = 0.527 grams
molar mass of Na₃PO₄.12H₂O = 380.12 g/mol
molar mass of BaCl₂.2H₂O = 244.26 g/mol
Determine the number of grams of Ba₃(PO₄)₂ produced
mole of Na₃PO₄.12H₂O = mass / molar mass
= 0.629 / 380.12 = 0.001654 mol
molar mass of BaCl₂.2H₂O = 0.527 / 244.26 = 0.002157 mol
After mixture : Determine the limiting reactant by lower mole ratio
mole ratio of BaCl₂.2H₂O = 0.002157 / 3
= 0.0007191
mole ratio of Na₃PO₄.12H₂O = 0.001654 / 2
= 0.000827
therefore the limiting reactant is : BaCl₂.2H₂O
Final step : Determine the mass of BaCl₂.2H₂O
molar mass of BaCl₂.2H₂O = 601.93 g/mol
Therefore :
mass of BaCl₂.2H₂O = moles * molar mass
= 0.0007191 * 601.93
= 0.4328 gram
Hence we can conclude that The number of grams of Ba₃(PO₄)₂ that is produced is : 0.4328 gram
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