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Mice21 [21]
2 years ago
10

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
Chemistry
1 answer:
Mama L [17]2 years ago
7 0

The barium sulfate formed is 1.57 grams

The chemical reaction as

BaCl2 + Na2SO4 → 2NaCl + BaS04

The given date is

35 ml of 0.160 barium chloride

58 ml of 0.065 m sodium sulfate

35 ml of barium chloride = 0.0350 liters

58 ml of sodium sulfate = 0.050 liters

moles of barium chloride = 0.0160 / 0.0350

                                          = 0.4571 mol

moles of sodium sulfate = 0.065 / 0.058

                                        = 1.120

Barium sulfate =  moles of barium chloride + sodium sulfate

                        = 0.4571 + 1.12

                        = 1.57 gram

Hence the gram of barium sulfate formed is 1.57 gram

Learn more about the moles on

brainly.com/question/14783710

#SPJ4

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Calculate the grams of ethane present in a sample containing 0.4271 moles if the molar mass of ethane is 30.067 g/mol.
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The grams of ethane present in a sample containing 0.4271 mole is 12.84 g

<h3>Description of mole </h3>

The mole of a substance is related to it's mass and molar mass according to the following equation

Mole = mass / molar mass

With the above formula, we can obtain the mass of ethane. Details below

<h3>How to determine the mass of ethane</h3>

The following data were obtained from the question:

  • Mole of ethane = 0.4271 mole
  • Molar mass of ethane = 30.067 g/mol
  • Mass of ethane =?

The mass of ethane present in the sample can be obtained as follow:

Mole = mass / molar mass

Cross multiply

Mass = mole × molar mass

Mass of ethane = 0.4271 × 30.067

Mass of ethane = 12.84 g

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