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3 years ago

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Which contains the greatest mass of oxygen: 0.75 moles of ethanol ( C2H5OH ), 0.60 mole of formic acid ( HCO2H ), or 1.0 mole of

water ( H2O )? Why?

2 answers:

Luda [366]3 years ago

7 0

Answer:

In conclussion, 0.60 moles of HCOOH contains the greatest mass of O

Explanation:

Let's make some rules of three, to solve this problem:

1 mol of ethanol has 2 moles of C, 6 moles of H, and 1 mol of oxygen

Therefore, 0.75 moles of ethanol must have 0.75 mol of oyxgen

Let's convert the moles to mass → 0.75 mol . 16 g/ 1 mol = 12 g

1 mol of formic acid has 2 moles of H, 1 mol of C and 2 mol of oxygen

0.60 moles of formic acid must have (0.6 .2) / 1 = 1.2 mol of O

If we convert the amount to mass → 1.2 mol . 16 g/ 1mol = 19.2 g

1 mol of water has 1 mol of oyxgen

Therefore, we have 1 mol of oxygen with a mass of 16 g.

In conclussion, 0.60 moles of HCOOH contains the greatest mass of O

TiliK225 [7]3 years ago

6 0

Answer:

0.60 mole of formic acid contains the greatest mass of oxygen

(19.2 grams)

Explanation:

Step 1: Data given

Moles of ethanol = 0.75 moles

Molar mass ethanol = 46.07 g/mol

Moles of formic acid = 0.60 moles

Molar mass of formic acid = 46.03 g/mol

Moles of H2O = 1.0

Molar mass of H2O = 18.02 g/mol

Step 2: Calculate moles of oxygen in each

Ethanol: For 1 mol ethanol we have 1 mol oxygen

For 0.75 moles ethanol we have 0.75 moles O

Mass O = 0.75 moles * 16.0 g/mol = 12.0 grams O

Formic acid: For 1 mol formic acid, we have 2 moles O

For 0.60 moles formic acid, we have 2*0.60 = 1.20 moles O

Mass O = 1.20 moles * 16.0 g/mol = 19.2 grams O

H2O: for 1.0 mol H2O we have 1 mol O

Mass O = 18.02 g/mol * 1.0 mol = 18.02 grams

0.60 mole of formic acid contains the greatest mass of oxygen

(19.2 grams)

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3 years ago

Read 2 more answers

4.33 g of 3-hexanol were obtained from 5.84 g of hex-3-ene. Determine the percentage yield of 3-hexanol. a Determine the moles o

Vilka [71]

<u>Answer:</u> The amount of hex-3-ene used is 0.0711 moles and the percent yield of 3-hexanol is 59.56 %

<u>Explanation:</u>

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

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

Given mass of hex-3-ene = 5.84 g

Molar mass of hex-3-ene = 82.14 g/mol

Putting values in equation 1, we get:

$\text{Moles of hex-3-ene}=\frac{5.84g}{82.14/mol}=0.0711mol$

The chemical equation for the conversion of hex-3-ene to 3-hexanol follows:

$\text{hex-3-ene}+H_2O\xrightarrow []{10\% H_2SO_4} \text{3-hexanol}$

By Stoichiometry of the reaction:

1 mole of hex-3-ene produces 1 mole of 3-hexanol

So, 0.0711 moles of hex-3-ene will produce = $\frac{1}{1}\times 0.0711=0.0711mol$ of 3-hexanol

Now, calculating the mass of 3-hexanol from equation 1, we get:

Molar mass of 3-hexanol = 102.2 g/mol

Moles of 3-hexanol = 0.0711 moles

Putting values in equation 1, we get:

$0.0711mol=\frac{\text{Mass of 3-hexanol}}{102.2g/mol}\\\\\text{Mass of 3-hexanol}=(0.0711mol\times 102.2g/mol)=7.27g$

To calculate the percentage yield of 3-hexanol, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of 3-hexanol = 4.33 g

Theoretical yield of 3-hexanol = 7.27 g

Putting values in above equation, we get:

$\%\text{ yield of 3-hexanol}=\frac{4.33g}{7.27g}\times 100\\\\\% \text{yield of 3-hexanol}=59.56\%$

Hence, the amount of hex-3-ene used is 0.0711 moles and the percent yield of 3-hexanol is 59.56 %

4 0

3 years ago

Convert 3 mol of Co, to grams.

GuDViN [60]

Answer:

176.7996 grams

Explanation:

Molar mass of Cobalt is 58.9332g. 58.9332 *3 is 176.7996g.

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2 years ago

Given the following data:

bagirrra123 [75]

$176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its $\Delta H$ can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with $\Delta H$ given be denoted as (1), (2), (3), and the last equation (4). Let $a$ , $b$ , and $c$ be letters such that $a \times (1) + b \times (2) + c \times (3) = (4)$ . This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, $3 + (-1) = 2$ shall resemble the number of $\text{H}_2$ left on the product side when the second equation is directly added to the third. Similarly

$\text{NH}_4 \text{Cl} \; (s)$ : $-2 \; a = 1$
$\text{NH}_3\; (g)$ : $-2 \; b = -1$
$\text{HCl} \; (g)$ : $2 \; c = -1$

Thus

$a = -1/2\\b = 1/2\\c = -1/2$ and

$-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)$

Verify this conclusion against a fourth species involved- $\text{N}_2 \; (g)$ for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

$a + b = -1/2 + 1/2 = 0$

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

$\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

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3 years ago

How does thermal energy flow?

mart [117]

My answer is....A

sorry if im wrong

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3 years ago