Question

How many grams of barium sulfate, baso4, are produced if 25.34 ml of 0.113 m bacl2 completely react given the reaction: bacl2 + na2so4 → baso4 + 2 nacl?

Answer 1

Answer is: mass of barium sulfate is 0.668 grams.
Chemical reaction: BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl.
V(BaCl₂) = 25.34 mL ÷ 1000 mL/L = 0.02534 L.
c(BaCl₂) = 0.113 mol/L.
n(BaCl₂) = V(BaCl₂) · c(BaCl₂).
n(BaCl₂) = 0.02534 L · 0.113 mol/L.
n(BaCl₂) = 0.00286 mol.
From chemical reaction: n(BaCl₂) : n(BaSO₄) = 1 : 1.
n(BaSO₄) = 0.00286 mol.
m(BaSO₄) = n(BaSO₄) · M(BaSO₄).
m(BaSO₄) = 0.00286 mol · 233.4 g/mol.
m(BaSO₄) = 0.668 g.

Answer 2

Answer:

0.668 g of barium sulfate

Explanation:

Given,

Balanced chemical equation: BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl.

Volume of BaCl₂ = 25.34 mL x $\frac{1L}{1000 ml }$= 0.02534 L.

Molarity of BaCl₂ = 0.113 M

Molarilty = $\frac{moles of solute}{L of the solution }$

Moles of solute = Molarilty x L of the solution

Moles of BaCl₂ = 0.113 M x 0.02534 L = 0.00286 mol.

From the balanced chemical equation there is a 1:1 molar ratio between BaCl₂ and BaSO₄

Therefore, moles of BaCl₂ = moles of BaSO₄

Moles of BaSO₄ = 0.00286 mol.

Mass of BaSO₄ = moles of BaSO₄ x Molar mass of BaSO₄

Mass of BaSO₄ = 0.00286 mol x 233.4 g/mol.

Mass of BaSO₄ = 0.668 g.