Question

What mass of barium sulfate can be produced by
complete reaction of 27.4 g of Na2SO4?

Answer 1

`\large \boxed{\text{45.0 g}}`

Step-by-step 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

`\text{Moles of Na$_(2)$SO$_(4)$} = \text{27.4 g Na$_(2)$SO$_(4)$}* \frac{\text{1 mol Na$_(2)$SO$_(4)$}}{\text{142.04 g Na$_(2)$SO$_(4)$}}= \text{0.1929 mol Na$_(2)$SO$_(4)$}`

(b) Moles of BaSO₄

`\text{Moles of BaSO$_(4)$} = \text{0.1929 mol Na$_(2)$SO$_(4)$} * \frac{\text{1 mol BaSO$_(4)$}}{\text{1 mol Na$_(2)$SO$_(4)$}} = \text{0.1929 mol BaSO$_(4)$}`

(c) Mass of BaSO₄

`\text{Mass of BaSO$_(4)$} =\text{0.1929 mol BaSO$_(4)$} * \frac{\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$_(4)$}`