Question

3. What is the mass in grams of precipitate formed when 250.0 mL of 0.300 M ZnCl? is mixed with excess KOH in the following chemical reaction? ZnCl? (aq) + 2 KOH (aq) ? Zn(OH)? (s) + 2 KCl (aq) 4. What is the mass in grams of H? that can be formed from 53.6 grams of NH? in the following reaction? 2 NH?(g) ? 3 H?(g) + N?(g)

          3. What is the mass in grams of precipitate formed when 250.0 mL of 0.300 M ZnCl? is mixed with excess KOH in the following chemical reaction? ZnCl? (aq) + 2 KOH (aq) ? Zn(OH)? (s) + 2 KCl (aq)
4. What is the mass in grams of H? that can be formed from 53.6 grams of NH? in the following reaction? 2 NH?(g) ? 3 H?(g) + N?(g)

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3. What is the mass in grams of precipitate formed when 250.0 mL of 0.300 M ZnCl? is mixed with excess KOH in the following chemical reaction? ZnCl? (aq) + 2 KOH (aq) ? Zn(OH)? (s) + 2 KCl (aq)
4. What is the mass in grams of H? that can be formed from 53.6 grams of NH? in the following reaction? 2 NH?(g) ? 3 H?(g) + N?(g)

Added by Raquel B.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Ronald Prasad

08/18/2023

Step 1

First, we need to find the number of moles of ZnCl2. We can do this by multiplying the volume (in liters) by the molarity. 0.250 L * 0.300 mol/L = 0.075 mol ZnCl2 According to the balanced chemical equation, one mole of ZnCl2 produces one mole of Zn(OH)2. Show more…

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3. What is the mass in grams of precipitate formed when 250.0 mL of 0.300 M ZnCl2 is mixed with excess KOH in the following chemical reaction? ZnCl2 (aq) + 2 KOH (aq) → Zn(OH)2 (s) + 2 KCl (aq) 4. What is the mass in grams of H2 that can be formed from 53.6 grams of NH3 in the following reaction? 2 NH3 (g) → 3 H2 (g) + N2 (g)

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