Question

tu mélanges 200ml dune solution de nano3 à 0,5mol/l avec 500ml dune solution de caco3 à 0,5mol/l. lenseignant te fait alors réaliser que tu as utilisé un trop grand volume dune des deux solutions pour que les deux réactifs réagissent ensemble. il te demande alors didentifier la solution utilisée en trop grande quantité. 2nano3 + caco3 -> ca(no3)2 + h2o + co2

Explanation:

Step1: Calculate moles of \( NaNO_3 \)

Molarity \( M = \frac{n}{V} \), so \( n = M \times V \). Volume of \( NaNO_3 \) solution \( V_1 = 200\space ml = 0.2\space L \), molarity \( M_1 = 0.5\space mol/L \).
\( n_{NaNO_3} = 0.5\space mol/L \times 0.2\space L = 0.1\space mol \)

Step2: Calculate moles of \( CaCO_3 \)

Volume of \( CaCO_3 \) solution \( V_2 = 500\space ml = 0.5\space L \), molarity \( M_2 = 0.5\space mol/L \).
\( n_{CaCO_3} = 0.5\space mol/L \times 0.5\space L = 0.25\space mol \)

Step3: Use stoichiometry from reaction \( 2NaNO_3 + CaCO_3

ightarrow Ca(NO_3)_2 + H_2O + CO_2 \)
The mole ratio of \( NaNO_3 \) to \( CaCO_3 \) is \( 2:1 \). For \( n_{NaNO_3} = 0.1\space mol \), the required moles of \( CaCO_3 \) is \( \frac{0.1}{2} = 0.05\space mol \). We have \( 0.25\space mol \) of \( CaCO_3 \), which is more than required. For \( n_{CaCO_3} = 0.25\space mol \), required \( NaNO_3 \) is \( 2 \times 0.25 = 0.5\space mol \), but we have only \( 0.1\space mol \) of \( NaNO_3 \). So \( CaCO_3 \) is in excess? Wait, no—wait, the volume of \( CaCO_3 \) is 500ml, \( NaNO_3 \) is 200ml. Wait, recalculate: Wait, the reaction is \( 2NaNO_3 + CaCO_3 \). So moles of \( NaNO_3 \): 0.2L 0.5mol/L = 0.1mol. Moles of \( CaCO_3 \): 0.5L 0.5mol/L = 0.25mol. The stoichiometric ratio is \( NaNO_3 : CaCO_3 = 2:1 \). So for 0.1mol of \( NaNO_3 \), we need \( 0.1 / 2 = 0.05\space mol \) of \( CaCO_3 \). We have 0.25mol of \( CaCO_3 \), which is way more. But the volume of \( CaCO_3 \) solution is 500ml, \( NaNO_3 \) is 200ml. Wait, maybe I mixed up. Wait, the problem is to find which solution ( \( NaNO_3 \) or \( CaCO_3 \) ) was used in too large a volume. Let's find the volume required for each to react completely.

For \( NaNO_3 \) (0.1mol) to react completely, moles of \( CaCO_3 \) needed: 0.05mol. Volume of \( CaCO_3 \) solution needed: \( V = \frac{n}{M} = \frac{0.05\space mol}{0.5\space mol/L} = 0.1\space L = 100\space ml \). But we used 500ml of \( CaCO_3 \) solution.

For \( CaCO_3 \) (0.25mol) to react completely, moles of \( NaNO_3 \) needed: 0.5mol. Volume of \( NaNO_3 \) solution needed: \( V = \frac{0.5\space mol}{0.5\space mol/L} = 1\space L = 1000\space ml \). But we used only 200ml of \( NaNO_3 \) solution.

So the solution used in too large a quantity (volume) is \( CaCO_3 \) solution (since we used 500ml instead of 100ml needed for \( NaNO_3 \) to react, while \( NaNO_3 \) was used in less volume than needed for \( CaCO_3 \) to react). Wait, no—wait, the question is "la solution utilisée en trop grande quantité" (the solution used in too large a quantity). So we used 200ml of \( NaNO_3 \) and 500ml of \( CaCO_3 \). From the stoichiometry, to react with 200ml (0.1mol) of \( NaNO_3 \), we need only 100ml of \( CaCO_3 \) solution (since 0.05mol of \( CaCO_3 \) is in 100ml of 0.5mol/L solution). But we used 500ml of \( CaCO_3 \), which is 5 times more than needed. For \( CaCO_3 \) (500ml, 0.25mol) to react, we need 1000ml of \( NaNO_3 \) solution, but we used only 200ml, which is less than needed. So the solution used in excess volume is \( CaCO_3 \) solution.

Answer:

The solution used in too large a quantity is the \( CaCO_3 \) solution (or "solution de \( CaCO_3 \)").