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2 years ago

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

Chemistry

2 answers:

ikadub [295]2 years ago

6 0

Answer: The moles of copper (I) hydroxide are 0.70 moles.

Explanation:

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]2 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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When heat is applied to 80 grams of CaCO3, it yields 39 grams of B. Determine the percentage of the yield.

EleoNora [17]

The question is incomplete, the complete question is:

When heat is applied to 80 grams of CaCO3, it yields 39 grams of CaO Determine the percentage of the yield.

CaCO3→CaO + CO2

Answer: The % yield of the product is 87.05 %

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

We are given:

Given mass of $CaCO_3$ = 80 g

Molar mass of $CaCO_3$ = 100 g/mol

Putting values in equation 1, we get:

$\text{Moles of }CaCO_3=\frac{80g}{100g/mol}=0.8mol$

For the given chemical reaction:

$CaCO_3\rightarrow CaO+CO_2$

By stoichiometry of the reaction:

If 1 mole of $CaCO_3$ produces 1 mole of CaO

So, 0.8 moles of $CaCO_3$ will produce = $\frac{1}{1}\times 0.8=0.8mol$ of CaO

We know, molar mass of $CaO$ = 56 g/mol

Putting values in above equation, we get:

$\text{Mass of CaO}=(0.8mol\times 56g/mol)=44.8g$

The percent yield of a reaction is calculated by using an equation:

$\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100$ ......(2)

Given values:

Actual value of the product = 39 g

Theoretical value of the product = 44.8 g

Plugging values in equation 2:

$\% \text{yield}=\frac{39 g}{44.8g}\times 100\\\\\% \text{yield}=87.05\%$

Hence, the % yield of the product is 87.05 %

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2 years ago

An atom of gold has a mass of 3.271 x 10-22g. How many atoms of gold are in 5.00g of gold?

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