consider the following reaction: Mg(s)+ 2HCl(aq) --> MgCl2 (aq)+H2 (g)
What minimum amount of 1.75 M HCl is necessary to produce 27.5 L of H2 at STP?
See above.
To find the minimum amount of 1.75 M HCl necessary to produce 27.5 L of H2 at STP, we need to use stoichiometry.
First, let's determine the balanced chemical equation:
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
According to the balanced equation, for every 2 moles of HCl, we get 1 mole of H2.
To find the amount of HCl needed, we can use the following formula:
Amount of HCl (in moles) = Volume (in liters) × Concentration (in moles per liter)
Given that the volume of H2 is 27.5 L and the desired ratio is 1 mole of H2 to 2 moles of HCl, we can set up the following proportion:
27.5 L H2 / 1 = n mol HCl / 2
Solving for n, the amount of HCl in moles, we get:
n = (27.5 L H2 × 2) / 1
n = 55 moles of HCl
Now, we can calculate the minimum amount of 1.75 M HCl needed. The concentration is given as 1.75 moles per liter, so:
Amount of 1.75 M HCl (in liters) = Amount of HCl (in moles) / Concentration (in moles per liter)
Substituting the values, we get:
Amount of 1.75 M HCl = 55 moles / 1.75 mol/L
Amount of 1.75 M HCl = 31.43 L
Therefore, the minimum amount of 1.75 M HCl necessary to produce 27.5 L of H2 at STP is approximately 31.43 liters.