2HBr+Mg(OH)2 -> MgBr2+2H2O
how many mL of a 0.2 M HBr solution is required to react with 40 mL of a 0.40 M Mg(OH)2 solution
How many mols Mg(OH)2 do you have? That's mols = M x L = ?
How many mols HBr will that use. The coefficients tell you 2 mols HBr are required for every mol Mg(OH)2.
Then M HBr = mols HBr/L HBr. You know mols and M, substitute and solve for L, then convert to mL.
Post your work if you get stuck.
To determine the volume of a 0.2 M HBr solution required to react with 40 mL of a 0.40 M Mg(OH)2 solution, you need to use the stoichiometry of the balanced chemical equation.
First, let's write the balanced chemical equation for the reaction:
2 HBr + Mg(OH)2 -> MgBr2 + 2 H2O
From the equation, we can see that 2 moles of HBr react with 1 mole of Mg(OH)2. This means that the ratio of moles of HBr to moles of Mg(OH)2 is 2:1.
Now, let's calculate the number of moles of Mg(OH)2 present in the given volume:
moles of Mg(OH)2 = concentration of Mg(OH)2 × volume of Mg(OH)2 solution
= 0.40 mol/L × 0.040 L
= 0.016 mol
Since the ratio of moles of HBr to moles of Mg(OH)2 is 2:1, we will need twice the number of moles of HBr. Therefore:
moles of HBr required = 2 × moles of Mg(OH)2
= 2 × 0.016 mol
= 0.032 mol
Now, let's calculate the volume of the 0.2 M HBr solution needed to supply 0.032 moles of HBr:
volume of HBr solution = moles of HBr required / concentration of HBr solution
= 0.032 mol / 0.2 mol/L
= 0.16 L
Since the concentration is given in Molarity (mol/L), we need to convert the volume to milliliters:
volume of HBr solution = 0.16 L × 1000 mL/L
= 160 mL
Therefore, 160 mL of the 0.2 M HBr solution is required to react with 40 mL of the 0.40 M Mg(OH)2 solution.