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Sever21 [200]
4 years ago
6

One of relatively few reactions that takes place directly between two solids at room temperature is In this equation, the in Ba(

OH)2 indicates the presence of eight water molecules. This compound is called barium hydroxide octahydrate.
a. Balance the equation.
b. What mass of ammonium thiocyanate (NH4SCN) must be used if it is to react completely with 6.5 g barium hydroxide octahydrate?
Chemistry
1 answer:
yuradex [85]4 years ago
3 0

Answer:

3.14 grams of ammonium thiocyanate must be used to react completely with 6.5 g barium hydroxide octahydrate.

Explanation:

Ba(OH)_2.8H_2O(s)+NH_4SCN(s)\rightarrow Ba(SCN)_2(s)+8H_2O(l)+NH_3(g)

The balance chemical equation is :

Ba(OH)_2.8H_2O(s)+2NH_4SCN(s)\rightarrow Ba(SCN)_2(s)+10H_2O(l)+2NH_3(g)

Mass of barium hydroxide octahydrate = 6.5 g

Moles of  barium hydroxide octahydrate = \frac{6.5 g}{315 g/mol}=0.020635 mol

According to reaction, 2 moles of ammonium thiocyanate reacts with1 mole of  barium hydroxide octahydrate. The 0.020635 moles of barium hydroxide octahydrate will react with:

\frac{2}{1}\times 0.020635 mol=0.04127 mol

Mass of 0.04127 moles of ammonium thiocyanate;

0.04127 mol\times 76 g/mol=3.136 g\approx 3.14 g

3.14 grams of ammonium thiocyanate must be used to react completely with 6.5 g barium hydroxide octahydrate

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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>

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<em>ksp = [borate²⁻] [Na⁺]²</em>

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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!

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