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?

Question

2 years ago

10

do not list units in your answer. stick to three sig figs

1 answer:

Mama L [17] · Answer 1 · 2023-06-02T23:37:11+03:00

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

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