Answer:
0.109 g.
Explanation:
Equation of the reaction:
Na3PO4 + 3HCl --> 3NaCl + H3PO4
Number of moles of HCl = molar concentration × volume
= 0.1 × 0.04
= 0.004 mol.
By stoichiometry, 1 mole of Na3PO4 neutralises 3 moles of HCl. Therefore, number of moles of Na3PO4 = 0.004/3
= 0.0013 mol
Mass of Na3PO4 = molar mass × number of moles
= 0.0013 × 164
= 0.219 g
Since 50% of Na3PO4 was present in the sample. Let 100 g be the total mass of the substance
= 0.219 × 50 g/100 g
= 0.109 g.
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You have to find the stoichiometric ratio between AlCl₃ and BaCl₂. The common element between them is Cl. So, the ratio of Cl in BaCl₂ to AlCl₃ is 2/3. The molar mass of AlCl₃ is 133.34 g/mol. The solution is as follows:
Mass of AlCl₃ = (6 mol BaCl₂)(2 mol Cl/1 mol BaCl₂)(1 mol AlCl₃/3 mol Cl)(133.34 g/mol) = 533.36 g AlCl₃