Answer :
To determine the number of ions from 63.4 grams of Mn(OH)2, calculate the moles using the molar mass then multiply by stoichiometric coefficients to find moles of Mn^2+ and OH^- ions produced. For this amount, we get 0.713 moles of Mn^2+ ions and 1.426 moles of OH^- ions.
To find out how many ions can be obtained from 63.4 grams of Mn(OH)2, we should first understand that Mn(OH)2, which stands for manganese(II) hydroxide, dissociates into ions when dissolved in water, releasing Mn^2+ and two OH^- for every formula unit that dissolves.
To convert grams of Mn(OH)2 to moles, we use the formula:
moles = mass (in grams) / molar mass (in grams per mole)
The molar mass of Mn(OH)2 can be calculated as:
Mn = 54.94 g/mol (1 Mn atom)
OH = 17.01 g/mol (2 O and 2 H atoms combined)
Molar mass of Mn(OH)2 = 54.94 + 2(17.01) = 88.96 g/mol
Now, calculate the moles of Mn(OH)2:
moles = 63.4 g / 88.96 g/mol = approximately 0.713 moles
Since each mole of Mn(OH)2 produces 1 mole of Mn^2+ ions and 2 moles of OH^- ions upon dissolution, we multiply by the respective stoichiometric coefficients to find the total number of ions produced:
Mn^2+ ions = 0.713 moles
OH^- ions = 2 * 0.713 moles = 1.426 moles
In summary, 63.4 grams of Mn(OH)2 will yield 0.713 moles of Mn^2+ ions and 1.426 moles of OH^- ions when dissolved in water.
