One mole of which hydrocarbon requires 8 mol o2 to achieve complete combustion to give carbon dioxide and water?

1 answer:

7 0

The typical structure of a hydrocarbon is HCnHnH where for every carbon atom, there's 2 hydrogen atoms plus a single hydrogen atom at each end of the hydrocarbon chain. So the expected output will be 1 CO2 molecule and 1 H2O molecule per carbon atom plus 1 extra H2O molecule. These molecules have to have a total of 16 oxygen atoms. So 2n + (n+1) = 16 where n = number of carbon atoms Solving for n gives 3n + 1 = 16 3n = 15 n = 5 So the equation should be: C5H7 + 8O2 ==> 5CO2 + 6H2O

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How many grams of hydrogen chloride can be produced from 1.00g of hydrogen and 55.0g of chlorine? what is the limiting reactant?

zloy xaker [14]

Answer:

$m_{HCl}=36.1gHCl$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the required grams of HCl by firstly identifying the limiting reactant via the moles of each reactant as they are in a 1:1 mole ratio:

$n_{H_2}=1.00gH_2*\frac{1molH_2}{2.02gH_2}=0.500molH_2\\\\ n_{Cl_2}=55.0gCl_2*\frac{1molCl_2}{70.9gCl_2}=0.776molCl_2$

Thus, we infer the hydrogen is the limiting reactant and therefore we use its 1:2 mole ratio with HCl whose molar mass is 36.46 g/mol:

$m_{HCl}=0.500molH_2*\frac{2molHCl}{1molH_2}*\frac{36.46gHCl}{1molHCl}\\\\m_{HCl}=36.1gHCl$

Regards!

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Nitration of 1,4−dimethylbenzene (p−xylene) gives a single product having the molecular formula c8h9no2 in high yield. draw this

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Structure is in document below.
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