How many liters of 0.250 M HCl would be required to react completely with 7.50 grams of magnesium hydroxide?
Mg(OH)2 + 2HCl ==> MgCl2 + 2H2O
Follow the steps is this example stoichiometry problem. You get moles by moles = M x L instead of step 1 as shown.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the number of liters of 0.250 M HCl required to react completely with 7.50 grams of magnesium hydroxide, you need to use stoichiometry and the concept of molar ratios.
First, let's write the balanced chemical equation for the reaction between magnesium hydroxide (Mg(OH)2) and hydrochloric acid (HCl):
Mg(OH)2 + 2HCl -> MgCl2 + 2H2O
From the balanced equation, we can see that 1 mole of Mg(OH)2 reacts with 2 moles of HCl.
Next, we need to convert the mass of Mg(OH)2 to moles. The molar mass of Mg(OH)2 is calculated as:
24.31 g/mol (Molar mass of Mg) + 2 * 16.00 g/mol (Molar mass of O) + 2 * 1.01 g/mol (Molar mass of H) = 58.33 g/mol
Using the molar mass, we can calculate the number of moles of Mg(OH)2:
7.50 g Mg(OH)2 * (1 mol Mg(OH)2 / 58.33 g Mg(OH)2) = 0.1283 mol Mg(OH)2
According to the balanced equation, we know that 1 mol Mg(OH)2 reacts with 2 moles of HCl. Therefore, 0.1283 mol Mg(OH)2 will react with:
0.1283 mol Mg(OH)2 * (2 mol HCl / 1 mol Mg(OH)2) = 0.2566 mol HCl
Now, we need to convert the moles of HCl to liters using the molarity (0.250 M). The molarity is defined as moles of solute per liter of solution. Therefore:
0.2566 mol HCl * (1 L / 0.250 mol HCl) = 1.0264 L
So, 1.0264 liters of 0.250 M HCl would be required to react completely with 7.50 grams of magnesium hydroxide.