Question

HELP. I DONT KNOW WHAT TO DO. Barium sulfate, BaSO4, is made by the following reaction.

Ba(NO3)2(aq) + Na2SO4(aq)
BaSO4(s) + 2NaNO3(aq)
What is the theoretical yield of barium sulfate, expressed in grams, if 75.00 g of Ba(NO3)2 were used with an excess of Na2SO4?
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Answer 1

`\boxed{\text{66.95 g BaSO$_(4)$}}`

Step-by-step explanation:

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.

M_r: 261.34 233.39

Ba(NO₃)₂ + Na₂SO₄ ⟶ BaSO₄ + 2NaNO₃

m/g: 75.00

1. Moles of Ba(NO₃)₂

`\text{Moles of Ba(NO$_(3))_(2)$} = \text{75.00 g} * \frac{\text{1 mol}}{\text{261.34 g}} = \text{0.286 98 mol}`

2. Moles of BaSO₄

The molar ratio is (1 mol BaSO₄/1 mol Ba(NO₃)₂

`\text{Moles of BaSO$_(4)$}= \text{0.286 98 mol Ba(NO$_(3))_(2)$ } * \frac{\text{1 mol BaSO$_(4)$}}{\text{1 mol Ba(NO$_(3))_(2)$}} = \text{0.286 98 mol BaSO$_(4)$}`

3. Mass of BaSO₄

`\text{Mass of BaSO$_(4)$} = \text{0.286 98 mol BaSO$_(4)$} * \frac{\text{233.39 g BaSO$_(4)$}}{\text{1 mol BaSO$_(4)$}} = \textbf{66.98 g BaSO$_(4)$}\\\\\text{The theoretical yield of barium sulfate is } \boxed{\textbf{66.98 g BaSO$_(4)$}}`