chemistry question help

Magnesium reacts with hydrochloric acid to make magnesium chloride and hydrogen. Calculate the volume of 1.00moldm-3 hydrochloric acid required to react with exactly 10.0mg of magnesium. Give your answer to an appropriate number of significant figures.

Reply 1

ricecakes1

find the number of moles for Mg = 0.01/24.3 = 4.12x10^-4 then do molar ratio so HCl has same number of moles as Mg. then use c=n/v formula and calculate the volume of HCl

Reply 2

Nitrotoluene

Original post by annoymous72

To calculate the volume of 1.00 mol/dm^3 of hydrochloric acid (HCl) required to react with 10.0 mg of magnesium (Mg), we must follow the steps below:
1: Write the balanced chemical equation
The reaction between magnesium and HCl is:

Mg + 2 HCl ==> MgCl2 + H2

2: Calculate the moles of magnesium
1. Convert the mass of magnesium into grams:

10.0 mg = 0.0100 g

2. Calculate the moles of magnesium using the molar mass of magnesium (about 24.31 g/mol):

Moles of Mg = mass of Mg/molar mass of Mg = 0.0100 g/24.31 g/mol = approx. 0.000411 mol

3: Determine the moles of hydrochloric acid required
From the balanced equation we can see that 1 mol of Mg reacts with 2 mol of HCl. Therefore
Required moles of HCl = 2 x moles of Mg = 2 x 0.000411 mol = about 0.000822 mol
4: Calculate the amount of hydrochloric acid required
Using the concentration of the hydrochloric acid
Volume (dm^3) = moles of HCl / concentration of HCl (mol/dm^3) = 0.000822 mol/1.00 mol/dm^3 = 0.000822 dm^3
5: Convert volume from dm^3 to mL
To convert from dm^ to mL
0.000822 dm^3 = 0.822 mL
Conclusion
The volume of 1.00 mol/dm^3 hydrochloric acid required to react with exactly 10.0 mg of magnesium is 0.822 mL (to three significant figures).