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3 years ago

11

Write the complete balanced equation for the following reaction:

1 answer:

hodyreva [135]3 years ago

8 0

To complete the balancing of the following combustion reaction, we must do elemental balances. For C, there is 5 on the left side hence there should be 5 CO2. For H, H20 should be 9/2. Next we balance O. on theright side we find a total of 29/2. So the o2 on the left side should be 29/4. To have whole number stoich.coeff. we multiply the numbers by 4, hence answer is
4C5H9O + 9O2= 20 O2 + 18 H20.

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never [62]

In order to find molarity, you must first find the number of moles that was dissolved.

Now, Moles = Mass ÷ Molar Mass

⇒ Moles of NaCl = 2.922 g ÷ 58.44 g/mol

= 0.05 moles

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3 years ago

If you have a solution that contains 46.85 g of codeine, C18H21NO3, in 125.5 g of ethanol, C2H5OH, what is the mole fraction of

irakobra [83]

Answer:

0.22

Explanation:

Given, Mass of $C_{18}H_{21}NO_3$ = 46.85 g

Molar mass of $C_{18}H_{21}NO_3$ = 299.4 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{46.85\ g}{299.4\ g/mol}$

$Moles\ of\ C_{18}H_{21}NO_3= 0.1565\ mol$

Given, Mass of $C_{2}H_{5}OH$ = 125.5 g

Molar mass of $C_{2}H_{5}OH$ = 46.07 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{125.5\ g}{46.07\ g/mol}$

$Moles\ of\ C_{2}H_{5}OH= 0.5535\ mol$

So, according to definition of mole fraction:

$Mole\ fraction\ of\ codeine=\frac {n_{codeine}}{n_{codeine}+n_{ethanol}}$

$Mole\ fraction\ of\ codeine=\frac{0.1565}{0.1565+0.5535}=0.22$

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Explanation:

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