Calculate the volume of 0.07216 M AgNO3 needed to react exactly with 0.3572 g of pure Na2CO3 to produce solid Ag2CO3.

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Chemistry Ngọc Hoa 4 years 2021-07-31T05:37:54+00:00 2021-07-31T05:37:54+00:00 1 Answers 52 views 0

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Captcha* Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )

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dangkhoa · Answer 1 · 2021-07-31T05:39:12+00:00

Answer:

93.4 mL

Explanation:

Let’s state the reaction:

2AgNO₃ + Na₂CO₃ → Ag₂CO₃ + 2NaNO₃

We determine the moles of sodium carbonate:

0.3572 g . 1mol / 105.98g = 3.37×10⁻³ moles

Ratio is 1:2. We say:

1 mol of sodium carbonate react to 2 moles of silver nitrate

Then, our 3.37×10⁻³ moles of carbonate may react to: 3.37×10⁻³ . 2

= 6.74×10⁻³ moles

If we convert to mmoles → 6.74×10⁻³ mol . 1000 mmol / mol = 6.74 mmol

Molarity is mol/L but we can use mmol /mL

6.74 mol / volume in mL = 0.07216 M

6.74 mol / 0.07216 M = volume in mL → 93.4 mL