If 45.5 mL of 0.150 M sodium sulfate solution reacts completely with aqueous barium nitrate,what is the mass of BaSO₄ (233.40 g/mol)precipitate? Ba(NO₃)2(aq)+ Na₂SO₄(aq)→ BaSO₄(s)+ 2 NaNO₃(aq)

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The Answer of If 45.5 mL of 0.150 M sodium sulfate solution reacts completely with aqueous barium nitrate,what is the mass of BaSO₄ (233.40 g/mol)precipitate? Ba(NO₃)2(aq)+ Na₂SO₄(aq)→ BaSO₄(s)+ 2 NaNO₃(aq)

If 455 ML of 0