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kondaur [170]
2 years ago
6

Activity A moles need hlp

Chemistry
1 answer:
damaskus [11]2 years ago
7 0
I would always start by balancing your carbons, and then balancing the rest from there.

1. C2H5OH + O2 —> CO2 + H2O - You have two carbons on the left and one on the right. Multiply CO2 by 2.
C2H5OH + O2 —> 2CO2 + H2O

Now balance hydrogen. You have 6 on the left and 2 on the right. Multiply H2O by 3.
C2H5OH + O2 —> 2CO2 + 3H2O

Now balance oxygen. You have 3 on the left and 7 on the right. You need 4 more on the left. Don’t multiply the C2H5OH by anything because that will change the numbers of everything else too. Multiply O2 by 3 instead.
C2H5OH + 3O2 —> 2CO2 + 3H2O

Check that all atoms are now balanced, and you’re good.

2. Same process as before.

First carbons - C3H8 + O2 —> 3CO2 + H2O
Then hydrogens - C3H8 + O2 —> 3CO2 + 4H2O
Then oxygens - C3H8 + 5O2 —> 3CO2 + 4H2O

3. Same again.

Carbons) C6H12O6 + O2 —> 6CO2 + H2O
Hydrogens) C6H12O6 + O2 —> 6CO2 + 6H2O
Oxygens) C6H12O6 + 6O2 —> 6CO2 + 6H2O

4. The general reaction for a combustion reaction is a hydrocarbon reacting with oxygen to produce carbon dioxide and water.
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Determine the percent water in MgSO4*7H20?
Semmy [17]

Answer:

\% H_2O=51.2\%

Explanation:

Hello!

In this case, since the percent water is computed by dividing the amount of water by the total mass of the hydrate; we infer we first need the molar mass of water and that of the hydrate as shown below:

MM_{MgSO_4* 7H_2O}=120.36 g/mol+7*18.02g/mol\\\\MM_{MgSO_4* 7H_2O}=246.5g/mol

Thus, the percent water is:

\% H_2O=\frac{7*MM_{H_2O}}{MM_{MgSO_4* 7H_2O}} *100\%\\\\

So we plug in to obtain:

\% H_2O=\frac{7*18.02}{246.5} *100\%\\\\\% H_2O=51.2\%

Best regards!

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