QuestionJune 17, 2025

When CuCl_(2) reacts with NaNO_(3) , NaCl and Cu(NO_(3))_(2) are produced as shown in the following balanced chemical equation. CuCl_(2)+2NaNO_(3)arrow 2NaCl+Cu(NO_(3))_(2) Suppose 19.35 g CuCl_(2) reacts with 20.31 NaNO_(3) Determine how many moles of Cu(NO_(3))_(2) will be produced if 19.35 g CuCl_(2) reacts with excess NaNO_(3) The molar mass of CuCl_(2) is 134.45g/mol

When CuCl_(2) reacts with NaNO_(3) , NaCl and Cu(NO_(3))_(2) are produced as shown in the following balanced chemical equation. CuCl_(2)+2NaNO_(3)arrow 2NaCl+Cu(NO_(3))_(2) Suppose 19.35 g CuCl_(2) reacts with 20.31 NaNO_(3) Determine how many moles of Cu(NO_(3))_(2) will be produced if 19.35 g CuCl_(2) reacts with excess NaNO_(3) The molar mass of CuCl_(2) is 134.45g/mol
When CuCl_(2) reacts with NaNO_(3) , NaCl and
Cu(NO_(3))_(2) are produced as shown in the following
balanced chemical equation.
CuCl_(2)+2NaNO_(3)arrow 2NaCl+Cu(NO_(3))_(2)
Suppose 19.35 g CuCl_(2) reacts with 20.31 NaNO_(3)
Determine how many moles of Cu(NO_(3))_(2) will be
produced if 19.35 g CuCl_(2) reacts with excess NaNO_(3)
The molar mass of CuCl_(2) is 134.45g/mol

Solution
4.2(369 votes)

Answer

0.144 \, \text{moles} Explanation 1. Calculate moles of CuCl_{2} Use the formula: \text{moles} = \frac{\text{mass}}{\text{molar mass}}. Moles of CuCl_{2} = \frac{19.35 \, \text{g}}{134.45 \, \text{g/mol}}. 2. Determine moles of Cu(NO_{3})_{2} produced From the balanced equation, 1 mole of CuCl_{2} produces 1 mole of Cu(NO_{3})_{2}. Therefore, moles of Cu(NO_{3})_{2} produced equals moles of CuCl_{2}.

Explanation

1. Calculate moles of $CuCl_{2}$<br /> Use the formula: $\text{moles} = \frac{\text{mass}}{\text{molar mass}}$. Moles of $CuCl_{2} = \frac{19.35 \, \text{g}}{134.45 \, \text{g/mol}}$.<br />2. Determine moles of $Cu(NO_{3})_{2}$ produced<br /> From the balanced equation, 1 mole of $CuCl_{2}$ produces 1 mole of $Cu(NO_{3})_{2}$. Therefore, moles of $Cu(NO_{3})_{2}$ produced equals moles of $CuCl_{2}$.
Click to rate:

Similar Questions