Answer
Verified
35.1k+ views
Hint: Try to recall that 1 mole of any substance at standard temperature and pressure (s.t.p) occupies volume of 22.4L. Now by using this you can easily answer the given question.
Complete step by step solution:
Given, mass of aluminium carbide, $A{l_4}{C_3}$ = 12g
Molar mass of aluminium carbide,$A{l_4}{C_3} = (27 \text{x} 4) + (12 \times 3)$
$\begin{gathered}
= 108 + 36
= 144{\text{ g/mol}}
\end{gathered}$
So, number of mole of aluminium carbide,$A{l_4}{C_3}$= $\dfrac{{{\text{mass of aluminium carbide}}}}{{molar{\text{ }}mass{\text{ of aluminium carbide}}}}$
$ = \dfrac{{12}}{{144}}$
$ = 0.083$.
In reaction, $A{l_4}{C_3} + 12{H_2}O \to 4Al{\left( {OH} \right)_3} + 3C{H_4}$:
1 mole of aluminium carbide,$A{l_4}{C_3}$ on reaction with water to form 4 moles of aluminium hydroxide, $Al{\left( {OH} \right)_3}$.
So, 0.083 mole of aluminium carbide,$A{l_4}{C_3}$ on reaction with water forms $ = 0.083 \times 4 = 0.33$ mole of aluminium hydroxide, $Al{\left( {OH} \right)_3}$.
Also, molar mass of aluminium hydroxide, $Al{\left( {OH} \right)_3}$ $ = (27) + 3 \times (16 + 1)$
$\begin{gathered}
= 27 + 51
= 78{\text{ g/mol}}
\end{gathered} $
Hence, mass of aluminium hydroxide, $Al{\left( {OH} \right)_3}$ formed = $78 \times 0.33$
= 25.74 g.
From first part you got, number of moles of aluminium carbide,$A{l_4}{C_3}$= 0.083
In reaction, $A{l_4}{C_3} + 12{H_2}O \to 4Al{\left( {OH} \right)_3} + 3C{H_4}$:
1 mole of aluminium carbide,$A{l_4}{C_3}$on reaction with water forms 3 mole of methane, $C{H_4}$.
So, 0.083 mole of aluminium carbide,$A{l_4}{C_3}$ on reaction with water forms $ = 0.083 \times 4 = 2.49$ mole of methane, $C{H_4}$.
Also, it is known to you that 1 mole of any substance occupies 22.4L of volume at standard temperature and pressure i.e. s.t.p.
Hence, 2.49 mole of methane, $C{H_4}$ occupies $ = 2.49 \times 22.4 = 55.7$ L of volume at s.t.p.
Therefore, from above we conclude that it forms 25.74 g mass of aluminium hydroxide, $Al{\left( {OH} \right)_3}$ and 55.77 L of methane at s.t.p.
Note:It should be remembered to you that aluminium carbide is used as an abrasive in cutting tools.
Also, you should remember that when magnesium carbide ($M{g_2}C$) reacts with water then it forms ethene gas and magnesium hydroxide.
Complete step by step solution:
Given, mass of aluminium carbide, $A{l_4}{C_3}$ = 12g
Molar mass of aluminium carbide,$A{l_4}{C_3} = (27 \text{x} 4) + (12 \times 3)$
$\begin{gathered}
= 108 + 36
= 144{\text{ g/mol}}
\end{gathered}$
So, number of mole of aluminium carbide,$A{l_4}{C_3}$= $\dfrac{{{\text{mass of aluminium carbide}}}}{{molar{\text{ }}mass{\text{ of aluminium carbide}}}}$
$ = \dfrac{{12}}{{144}}$
$ = 0.083$.
In reaction, $A{l_4}{C_3} + 12{H_2}O \to 4Al{\left( {OH} \right)_3} + 3C{H_4}$:
1 mole of aluminium carbide,$A{l_4}{C_3}$ on reaction with water to form 4 moles of aluminium hydroxide, $Al{\left( {OH} \right)_3}$.
So, 0.083 mole of aluminium carbide,$A{l_4}{C_3}$ on reaction with water forms $ = 0.083 \times 4 = 0.33$ mole of aluminium hydroxide, $Al{\left( {OH} \right)_3}$.
Also, molar mass of aluminium hydroxide, $Al{\left( {OH} \right)_3}$ $ = (27) + 3 \times (16 + 1)$
$\begin{gathered}
= 27 + 51
= 78{\text{ g/mol}}
\end{gathered} $
Hence, mass of aluminium hydroxide, $Al{\left( {OH} \right)_3}$ formed = $78 \times 0.33$
= 25.74 g.
From first part you got, number of moles of aluminium carbide,$A{l_4}{C_3}$= 0.083
In reaction, $A{l_4}{C_3} + 12{H_2}O \to 4Al{\left( {OH} \right)_3} + 3C{H_4}$:
1 mole of aluminium carbide,$A{l_4}{C_3}$on reaction with water forms 3 mole of methane, $C{H_4}$.
So, 0.083 mole of aluminium carbide,$A{l_4}{C_3}$ on reaction with water forms $ = 0.083 \times 4 = 2.49$ mole of methane, $C{H_4}$.
Also, it is known to you that 1 mole of any substance occupies 22.4L of volume at standard temperature and pressure i.e. s.t.p.
Hence, 2.49 mole of methane, $C{H_4}$ occupies $ = 2.49 \times 22.4 = 55.7$ L of volume at s.t.p.
Therefore, from above we conclude that it forms 25.74 g mass of aluminium hydroxide, $Al{\left( {OH} \right)_3}$ and 55.77 L of methane at s.t.p.
Note:It should be remembered to you that aluminium carbide is used as an abrasive in cutting tools.
Also, you should remember that when magnesium carbide ($M{g_2}C$) reacts with water then it forms ethene gas and magnesium hydroxide.
Recently Updated Pages
To get a maximum current in an external resistance class 1 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Let f be a twice differentiable such that fleft x rightfleft class 11 maths JEE_Main
Find the points of intersection of the tangents at class 11 maths JEE_Main
For the two circles x2+y216 and x2+y22y0 there isare class 11 maths JEE_Main
The path difference between two waves for constructive class 11 physics JEE_MAIN
Other Pages
when an object Is placed at a distance of 60 cm from class 12 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Which of the following sets of displacements might class 11 physics JEE_Main
Oxidation state of S in H2S2O8 is A 6 B 7 C +8 D 0 class 12 chemistry JEE_Main
What is the pH of 001 M solution of HCl a 1 b 10 c class 11 chemistry JEE_Main
The correct IUPAC name of the given compound is A isopropylbenzene class 11 chemistry JEE_Main