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Licemer1 [7]
3 years ago
6

A reaction produces 0.755 mol of H2O. How many molecules of water are produced?

Chemistry
1 answer:
Fynjy0 [20]3 years ago
7 0

Answer:

4.55 * 10^23 molecules

Explanation:

This is a conversion problem. We know that one mole has 6.022 *10^23 particles or molecules so we can use this as a conversion factor.

0.755 mol ( 6.022*10^23 molecules/ 1 mol)

=4.55*10^23 molecules

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Convert mass to moles for both reactants. (round to 2 significant figures.)
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Explanation:

Given parameters:

Mass of CuCl₂  = 2.50g

Mass of Al  = 0.50g

Unknown:

Number of moles of CuCl₂ and Al  = ?

Solution:

To solve this problem, we must understand that the number of moles is a fundamental property used in stoichiometry calculations.

         Number of moles  = \frac{mass}{molar mass}

Molar mass of CuCl₂  = 63.6 + 2(35.5) = 134.5g/mole

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3 years ago
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Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount requir
guajiro [1.7K]

Answer:

y_{O2} =4.3%

Explanation:

The ethanol combustion reaction is:

C_{2}H_{5} OH+3O_{2}→2CO_{2}+3H_{2}O

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}

Dividing the previous equation by x:

1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles

Calculate the number of moles of CO2 and water considering the same:

0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)

0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)

The total number of moles at the reactor output would be:

N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)

So, the oxygen mole fraction would be:

y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3%

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