
Calculate the exact number of grams of aluminum that should react with $ 35 $ mL of $ 2.0 $ M hydrochloric acid? Balance the given equation;
$ \_HCl + \_Al \to \_AlCl_3 + \_H_2 $
Answer
522.9k+ views
Hint: When we get an equation, we must always check if it is balanced. If it is not balanced, balance it. We use stoichiometry to do this. The procedure we must follow in order to reach the final answer is; first since molarity of hydrochloric acid is given, we must find the number of moles in it, then we can proceed to calculate the grams of aluminum that will satisfy the equation.
Complete answer:
Always start a question like this by writing down what all are given and what we need to find;
Molarity of hydrochloric acid ( $ HCl $ ) $ = 2.0M $
Volume of hydrochloric acid in liters ( $ HCl $ ) $ = 35mL = 0.035L $
To find: Number of grams of aluminum $ = ? $
Since our given equation is not balanced, by stoichiometry we can balance the equation. If the equation is balanced it is easier to calculate the quantities and we can only proceed to calculations after balancing any given unbalanced equation.
Balanced equation: $ 2Al + 6HCl \to 3H_2 + 2AlCl_3 $
Let us see what our next step is; it is to find for $ HCl $ the number of moles that are involved in reaction.
We use molarity for this:
$ Molarity = \dfrac{{number\;of\;moles\;of\;HCl}}{{volume\;of\;HCl}} $
$ \Rightarrow 2.0M = \dfrac{{number\;of\;moles}}{{0.035L}} $
$ \Rightarrow number\;of\;moles = 0.070 $ moles
$ \therefore $ $ 0.070 $ moles of $ HCl $ are needed in the reaction.
With the help of our balanced equation we can now determine number of grams of aluminum needed;
$ number\;of\;grams = (ratio\;of\;moles) \times (molecular\;mass\;of\;Al) \times (number\;of\;moles\;of\;HCl) $
$ number\;of\;grams = (\dfrac{{2mol}}{{6mol}})(\dfrac{{26.98g}}{{1mol}})(0.070\,moles) $
$ \therefore Number\;of\;grams = 0.63g $ of Aluminum is required to satisfy the equation.
Note:
Aluminum is a metal so it mostly reacts with non-metals, and since it exists at room temperature, in order for reactions to take place, external agents like heat must be used. Even though it is a metal it is proved that aluminum cannot rust, which is why it is very reliable. But usage of aluminum is not encouraged as it is difficult to degrade, that is it is non-biodegradable.
Complete answer:
Always start a question like this by writing down what all are given and what we need to find;
Molarity of hydrochloric acid ( $ HCl $ ) $ = 2.0M $
Volume of hydrochloric acid in liters ( $ HCl $ ) $ = 35mL = 0.035L $
To find: Number of grams of aluminum $ = ? $
Since our given equation is not balanced, by stoichiometry we can balance the equation. If the equation is balanced it is easier to calculate the quantities and we can only proceed to calculations after balancing any given unbalanced equation.
Balanced equation: $ 2Al + 6HCl \to 3H_2 + 2AlCl_3 $
Let us see what our next step is; it is to find for $ HCl $ the number of moles that are involved in reaction.
We use molarity for this:
$ Molarity = \dfrac{{number\;of\;moles\;of\;HCl}}{{volume\;of\;HCl}} $
$ \Rightarrow 2.0M = \dfrac{{number\;of\;moles}}{{0.035L}} $
$ \Rightarrow number\;of\;moles = 0.070 $ moles
$ \therefore $ $ 0.070 $ moles of $ HCl $ are needed in the reaction.
With the help of our balanced equation we can now determine number of grams of aluminum needed;
$ number\;of\;grams = (ratio\;of\;moles) \times (molecular\;mass\;of\;Al) \times (number\;of\;moles\;of\;HCl) $
$ number\;of\;grams = (\dfrac{{2mol}}{{6mol}})(\dfrac{{26.98g}}{{1mol}})(0.070\,moles) $
$ \therefore Number\;of\;grams = 0.63g $ of Aluminum is required to satisfy the equation.
Note:
Aluminum is a metal so it mostly reacts with non-metals, and since it exists at room temperature, in order for reactions to take place, external agents like heat must be used. Even though it is a metal it is proved that aluminum cannot rust, which is why it is very reliable. But usage of aluminum is not encouraged as it is difficult to degrade, that is it is non-biodegradable.
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