How many grams of barium sulfate form when 35.0 ml of 0.160 m barium chloride reacts with 58.0 ml of 0.065 m sodium sulfate? do not list units in your answer. stick to three sig figs.

Respuesta :

According to the given statement 0.88 grams of barium sulfate form when 35.0 ml of 0.160 m barium chloride reacts with 58.0 ml of 0.065 m sodium sulfate.

Mass:

A physical body's overall composition is measured by its mass.

Inertia, or the body's resistance to acceleration in the presence of a net force, is also measured by it.

The mass of a thing also affects how strongly it attracts other bodies through gravity.

In the SI, the kilogram serves as the basic mass unit (kg).

This information

Moles of BaCl2 equal 5.6 x 103 (35 x 0.160 / 1000)

sodium sulfate moles equal 58 x 0.065, or 3.77 x 103,

Na2SO4 with BaCl2 ——————————-> 2NaCl plus BaSO4

Na2SO4 is the limiting agent here.

Therefore, the product relies on limiting reagent.

1 mol of Na2SO4 produces 1 mol of BaSO4.

0.00377 moles of BaSO4 is one mole.

moles = mass / molar mass of BaSO4.

0.00377 = mass / 233.4

BaSO4 mass is 0.88 g.

So, 0.88 grams of barium sulfate make up one gram.

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