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3 years ago

How many moles are in 68 grams of copper (ll) hydroxide, Cu(OH)2

Chemistry

2 answers:

ikadub [295]3 years ago

6 0

<u>Answer:</u> The moles of copper (I) hydroxide are 0.70 moles.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

We are given:

Given mass of copper (I) hydroxide = 68 g

Molar mass of copper (I) hydroxide = 97.6 g/mol

Putting values in above equation, we get:

$\text{Moles of copper (I) hydroxide}=\frac{68g}{97.6g/mol}=0.70mol$

Hence, the moles of copper (I) hydroxide are 0.70 moles.

Sveta_85 [38]3 years ago

4 0

Molar mass Cu(OH)2 => 97.561 g/mol

Therefore:

1 mole Cu(OH)2 ------------ 97.561 g
? moles Cu(OH)2 ----------- 68 g

= 68 x 1 / 97.561

= 68 / 97.561

= 0.696 moles of Cu(OH)2

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Answer:

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As there are 62.5 moles of hydrogen:

<h3>Hydrogen is the excess reactant</h3><h3>Nitrogen is the limiting reactant</h3><h3 />

With nitrogen, the limiting reactant, we determine theoretical moles (Assuming 100% of the reaction occurs) and theoretical yield (In mass):

4.46 moles N2 * (2moles NH3 / 1mol N2) = 8.92 moles of ammonia

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8.92 moles of ammonia * (17g/mol) =

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