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max2010maxim [7]
1 year ago
12

1.562 g sample of the alcohol CH3CHOHCH2CH3 is burned in an excess of oxygen. What masses of H2O and CO2 should be obtained

Chemistry
1 answer:
Diano4ka-milaya [45]1 year ago
7 0

Answer:

m_{CO_2}=3.709gCO_2 \\\\m_{H_2O}=1.898gH_2O

Explanation:

Hello.

In this case, since the molecular formula of the given alcohol is C₄H₁₀O (molar mass = 74.14 g/mol), we can write its combustion reaction as shown below:

C_4H_1_0O+6O_2\rightarrow 4CO_2+5H_2O

Thus, since there is a 1:4 mole ratio with carbon dioxide (molar mass = 44..01 g/mol) and a 1:5 mole ratio with water (molar mass = 18.02 g/mol), we can compute the obtained masses as shown below:

m_{CO_2}=1.562gC_4H_1_0O*\frac{1mol}{74.14gC_4H_1_0O} *\frac{4molCO_2}{1molC_4H_1_0O} *\frac{44.01gCO_2}{1molCO_2}=3.709gCO_2 \\\\m_{H_2O}=1.562gC_4H_1_0O*\frac{1mol}{74.14gC_4H_1_0O} *\frac{5molH_2O}{1molC_4H_1_0O} *\frac{18.02gH_2O}{1molH_2O}=1.898gH_2O

Best regards!

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If the actual yield of a reaction is 37.6 g while the theoretical yield is 112.8 g what is the percent yield
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<h2>Hello!</h2>

The answer is:

The percent yield of the reaction is 32.45%

<h2>Why?</h2>

To calculate the percent yield, we have to consider the theoretical yield and the actual yield. The theoretical yield as its name says is the yield expected, however, many times the difference between the theoretical yield and the actual yield is notorious.

We are given that:

ActualYield=37.6g\\TheoreticalYield=112.8g

Now, to calculate the percent yield, we need to divide the actual yield by the theoretical and multiply it by 100.

So, calculating we have:

PercentYield=\frac{ActualYield}{TheoreticalYield}*100\\\\PercentYield=\frac{37.6g}{112.8g}*100=0.3245*100=32.45(percent)

Hence, we have that the percent yield of the reaction is 32.45%.

Have a nice day!

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So, let us check which is stable and which is unstable:

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h. 1 proton and 0 neutrons =  Neutron- proton ratio = N/P = 0/1 =0 = unstable

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