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Aleks04 [339]
3 years ago
10

What mass of Cu(s) would be produced if 0.40 mol of Cu2O(s) was reduced completely with excess H2(g) ?

Chemistry
1 answer:
yan [13]3 years ago
4 0

Answer:

50.8g

Explanation:

Given parameters:

Nubmer of mole of Cu₂O = 0.4mol

Unknown:

Mass of Cu produced = ?

Solution:

The balanced reaction equation:

           Cu₂O  +  H₂   ⇄ 2Cu + H₂O

From the reaction, we know that:

            1 mole of Cu₂O produced 2 moles of Cu

             0.4 mol of Cu₂O would give 0.8mol of Cu

Now we know the number of moles of Cu produced because the limiting reactant here is Cu₂O. It was used up in the reaction in which the hydrogen gas was in excess.

To find the mass of Cu produced, we use the equation below :

              Mass of Cu = Number of moles of Cu x molar mass of Cu

Molar mass of Cu = 63.5gmol⁻¹

              Mass of Cu = 0.8 x 63.5 = 50.8g

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A student who is performing this experiment pours an 8.50 mL sample of the saturated borax solution into a 10 mL graduated cylin
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Answer:

ksp = 0,176

Explanation:

The borax (Na₂borate) in water is in equilibrium, thus:

Na₂borate(s) ⇄ borate²⁻(aq) + 2Na⁺(aq)

<em>When you add just borax, the moles of Na²⁺ are twice the moles of borate²⁻, that means 2borate²⁻=Na⁺ </em><em>(1)</em>

The ksp is defined as:

<em>ksp = [borate²⁻] [Na⁺]²</em>

Then, borate²⁻(B₄O₇²⁻) reacts with HCl thus:

B₄O₇²⁻ + 2HCl + 5H₂O → 4H₃BO₃ + 2Cl⁻

The moles of HCl that reacts with B₄O₇²⁻ are:

0,500M×0,01200L = 6,00x10⁻³ mol of HCl

As two moles of HCl react with 1 mol of B₄O₇²⁻, the moles of B₄O₇²⁻ are:

6,00x10⁻³ mol of HCl×\frac{1molB_{4}O_{7}^{2-}}{2molHCl} = <em>3,00x10⁻³ mol of B₄O₇²⁻</em>

For (1), moles of Na⁺ are <em>3,00x10⁻³ mol ×2 = 6,00x10⁻³ mol of Na⁺</em>

The [borate²⁻] is <em>3,00x10⁻³ mol of B₄O₇²⁻/0,00850L = </em><em>0,353M</em>

And [Na⁺] is <em>6,00x10⁻³ mol of Na⁺ / 0,00850L = </em>0,706M

Replacing in the expression of ksp:

ksp = [0,353] [0,706]²

<em>ksp = 0,176</em>

<em></em>

I hope it helps!

8 0
3 years ago
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