Answer:
[tex]\large \boxed{\text{45.0 g}}[/tex]
Explanation:
We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.
MM: 142.04 233.29
Ba²⁺ + Na₂SO₄ ⟶ BaSO₄ + 2Na⁺
m/g: 27.4
(a) Moles of Na₂SO₄
[tex]\text{Moles of Na$_{2}$SO$_{4}$} = \text{27.4 g Na$_{2}$SO$_{4}$}\times \dfrac{\text{1 mol Na$_{2}$SO$_{4}$}}{\text{142.04 g Na$_{2}$SO$_{4}$}}= \text{0.1929 mol Na$_{2}$SO$_{4}$}[/tex]
(b) Moles of BaSO₄
[tex]\text{Moles of BaSO$_{4}$} = \text{0.1929 mol Na$_{2}$SO$_{4}$} \times \dfrac{\text{1 mol BaSO$_{4}$}}{\text{1 mol Na$_{2}$SO$_{4}$}} = \text{0.1929 mol BaSO$_{4}$}[/tex]
(c) Mass of BaSO₄
[tex]\text{Mass of BaSO$_{4}$} =\text{0.1929 mol BaSO$_{4}$} \times \dfrac{\text{233.29 g BaSO$_{4}$}}{\text{1 mol BaSO$_{4}$}} = \textbf{45.0 g BaSO$_{4}$}\\\\\text{The reaction produces $\large \boxed{\textbf{45.0 g}}$ of BaSO$_{4}$}[/tex]
