QuestionMay 27, 2025

Na_(2)SiO_(3)(s)+8HF(aq)... H_(2)SiF_(6)(aq)+2NaF(aq)+3H_(2)O(l) How many grams of excess reactant are left when 61g of Na_(2)SiO_(3) reacts with 60g of HF? Answer: square

Na_(2)SiO_(3)(s)+8HF(aq)... H_(2)SiF_(6)(aq)+2NaF(aq)+3H_(2)O(l) How many grams of excess reactant are left when 61g of Na_(2)SiO_(3) reacts with 60g of HF? Answer: square
Na_(2)SiO_(3)(s)+8HF(aq)... H_(2)SiF_(6)(aq)+2NaF(aq)+3H_(2)O(l) How many grams of excess reactant are left when 61g of Na_(2)SiO_(3) reacts with 60g of HF? Answer: square

Solution
4.2(184 votes)

Answer

0 \, \text{g} Explanation 1. Calculate moles of Na_{2}SiO_{3} Molar mass of Na_{2}SiO_{3} is 122.06 \, \text{g/mol}. Moles = \frac{61}{122.06} = 0.5 \, \text{mol}. 2. Calculate moles of HF Molar mass of HF is 20.01 \, \text{g/mol}. Moles = \frac{60}{20.01} = 3 \, \text{mol}. 3. Determine limiting reactant Reaction requires 8 moles of HF per mole of Na_{2}SiO_{3}. For 0.5 mol Na_{2}SiO_{3}, 4 mol HF needed. HF is excess. 4. Calculate excess HF Excess HF = Initial HF - Required HF = 3 - 4 = -1 \, \text{mol}. Since negative, all HF reacts; no excess HF. 5. Calculate grams of excess HF No excess HF, so 0 \, \text{g} excess.

Explanation

1. Calculate moles of $Na_{2}SiO_{3}$<br /> Molar mass of $Na_{2}SiO_{3}$ is $122.06 \, \text{g/mol}$. Moles = $\frac{61}{122.06} = 0.5 \, \text{mol}$.<br />2. Calculate moles of HF<br /> Molar mass of HF is $20.01 \, \text{g/mol}$. Moles = $\frac{60}{20.01} = 3 \, \text{mol}$.<br />3. Determine limiting reactant<br /> Reaction requires $8$ moles of HF per mole of $Na_{2}SiO_{3}$. For $0.5$ mol $Na_{2}SiO_{3}$, $4$ mol HF needed. HF is excess.<br />4. Calculate excess HF<br /> Excess HF = Initial HF - Required HF = $3 - 4 = -1 \, \text{mol}$. Since negative, all HF reacts; no excess HF.<br />5. Calculate grams of excess HF<br /> No excess HF, so $0 \, \text{g}$ excess.
Click to rate:

Similar Questions