MgCl2(aq) + 2 NaOH(aq) --> Mg(OH)2(s) + 2 NaCl(aq)
How many grams of Mg(OH)2 would form if 50.00 mL of 0.9750M NaOH were combined with 50.00 mL of 0.8820 M MgCl2?
To find the number of grams of Mg(OH)2 that would form, we need to use the balanced chemical equation and stoichiometry.
First, let's find the number of moles of NaOH and MgCl2 used in the reaction.
For NaOH:
- Given volume: 50.00 mL
- Convert to liters: 50.00 mL * (1 L / 1000 mL) = 0.05000 L
- Given molarity: 0.9750 M
- Calculate moles: moles = volume * molarity = 0.05000 L * 0.9750 M = 0.04875 moles
For MgCl2:
- Given volume: 50.00 mL
- Convert to liters: 50.00 mL * (1 L / 1000 mL) = 0.05000 L
- Given molarity: 0.8820 M
- Calculate moles: moles = volume * molarity = 0.05000 L * 0.8820 M = 0.04410 moles
Since the balanced chemical equation shows a 1:1 ratio between MgCl2 and Mg(OH)2, we know that the number of moles of Mg(OH)2 formed will be equal to the number of moles of MgCl2 used.
Therefore, the number of moles of Mg(OH)2 formed is 0.04410 moles.
To find the mass of Mg(OH)2 formed, we need to use its molar mass.
The molar mass of Mg(OH)2 is calculated as:
(1 * atomic mass of Mg) + (2 * atomic mass of O) + (2 * atomic mass of H)
- Atomic mass of Mg = 24.31 g/mol
- Atomic mass of O = 16.00 g/mol
- Atomic mass of H = 1.01 g/mol
Using these values, we can calculate the molar mass of Mg(OH)2:
(1 * 24.31 g/mol) + (2 * 16.00 g/mol) + (2 * 1.01 g/mol) = 58.33 g/mol
Finally, we can use the number of moles of Mg(OH)2 and its molar mass to calculate the mass of Mg(OH)2 formed.
Mass = moles * molar mass
Mass = 0.04410 moles * 58.33 g/mol
The mass of Mg(OH)2 formed would be approximately 2.571 grams.