91影视

The reaction between potassium hydroxide (KOH) and magnesium nitrate [Mg(NO鈧�)鈧俔 can be written as: KOH (aq) + Mg(NO鈧�)鈧� (aq) 鈫� ? To create a balanced chemical equation, we must first predict the products formed. According to the solubility rules, alkali metal ions (Group 1 elements) and nitrate ions are soluble, so we expect potassium nitrate (KNO鈧�) to be soluble. Magnesium hydroxide (Mg(OH)鈧�), on the other hand, is slightly soluble in water. Therefore, our balanced chemical equation should look like this: 2 KOH (aq) + Mg(NO鈧�)鈧� (aq) 鈫� 2 KNO鈧� (aq) + Mg(OH)鈧� (s)

In this reaction, the insoluble product formed is magnesium hydroxide (Mg(OH)鈧�). It forms as a solid precipitate, so we indicate this in our balanced chemical equation with the (s) state symbol: 2 KOH (aq) + Mg(NO鈧�)鈧� (aq) 鈫� 2 KNO鈧� (aq) + Mg(OH)鈧� (s)

To calculate the mass of Mg(OH)鈧� produced, we must identify the limiting reactant and use stoichiometry. We have 100.0 mL of 0.200 M KOH and 100.0 mL of 0.200 M Mg(NO鈧�)鈧� as our starting amounts. First convert the volumes to moles: moles KOH = volume 脳 concentration moles KOH = (100.0 mL)(0.200 mol/L) = 0.0200 mol moles Mg(NO鈧�)鈧� = volume 脳 concentration moles Mg(NO鈧�)鈧� = (100.0 mL)(0.200 mol/L) = 0.0200 mol Now, use stoichiometry to find the moles of Mg(OH)鈧� formed: moles Mg(OH)鈧� = moles Mg(NO鈧�)鈧� 脳 (1 mol Mg(OH)鈧� / 1 mol Mg(NO鈧�)鈧�) moles Mg(OH)鈧� = 0.0200 mol Mg(NO鈧�)鈧� 脳 (1 mol Mg(OH)鈧� / 1 mol Mg(NO鈧�)鈧�) = 0.0200 mol Mg(OH)鈧� Next, convert moles to mass: mass Mg(OH)鈧� = moles 脳 molar mass mass Mg(OH)鈧� = (0.0200 mol)(58.32 g/mol) = 1.1664 g Therefore, 1.1664 g of Mg(OH)鈧� (s) precipitate is produced.

Since Mg(OH)鈧� is the limiting reactant, 1 mole of Mg(NO鈧�)鈧� will react with 2 moles of KOH. After precipitation is complete, all moles of Mg(NO鈧�)鈧� will have reacted. However, we will have an excess of KOH remaining. Calculate the moles of KOH remaining: moles KOH remaining = moles KOH initially - (moles Mg(NO鈧�)鈧� 脳 2) moles KOH remaining = 0.0200 mol - (0.0200 mol 脳 2) = 0.0200 mol - 0.0400 mol = -0.0200 mol Since the remaining moles of KOH cannot be negative, the limiting reactant is KOH instead of Mg(NO鈧�)鈧�. moles KOH reacted = moles KOH initially / 2 moles reacted = 0.0200 mol / 2 = 0.0100 mol Now, find the moles of KNO鈧� formed and the moles of Mg(NO鈧�)鈧� remaining: moles KNO鈧� = moles KOH reacted 脳 2 moles KNO鈧� = 0.0100 mol 脳 2 = 0.0200 mol KNO鈧� moles Mg(NO鈧�)鈧� remaining = moles Mg(NO鈧�)鈧� initially - moles reacted moles Mg(NO鈧�)鈧� remaining = 0.0200 mol - 0.0100 mol = 0.0100 mol Finally, calculate the concentration of each ion in solution. The total volume of the solution is 200.0 mL. Concentration of K鈦�: \[c(K^+) = \frac{moles \, KNO_3}{total \, volume} = \frac{0.0200 \, mol \, KNO_3}{0.200 \, L} = 0.100 \, M\] Concentration of Mg虏鈦�: \[c(Mg^{2+}) = \frac{moles \, Mg(NO_3)_2}{total \, volume} = \frac{0.0100 \, mol \, Mg(NO_3)_2}{0.200 \, L} = 0.0500 \, M\] Concentration of NO鈧冣伝: \[c(NO_3^-) = \frac{moles \, Mg(NO_3)_2 \times 2 + moles \, KNO_3}{total \, volume} = \frac{(0.0100 \, mol \times 2) + 0.0200 \, mol}{0.200 \, L} = \frac{0.0400 \, mol}{0.200 \, L} = 0.200 \, M\] So, after precipitation is complete, the concentration of K鈦� is 0.100 M, Mg虏鈦� is 0.0500 M, and NO鈧冣伝 is 0.200 M.