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overall question: part 1 of 6 convert 92.4 grams of kf to grams of ag₃n…

Question

overall question:
part 1 of 6
convert 92.4 grams of kf to grams of ag₃n according to the balanced reaction below.
k₃n + 3agf → 3kf + ag₃n
part a:
what is the molar mass of kf?
round to two decimal places.
g/mol kf
what is the molar mass of ag₃n?
round to two decimal places.
g/mol ag₃n
submit part
question 13
0/4 pts 5 19 details
consider the following reaction:
2hcl + caco₃ → cacl₂ + h₂o + co₂
how many mols of calcium chloride can be produced if you begin with 3.65 ml of 0.62 m hcl and have an excess amount of calcium carbonate?

Explanation:

Part A (Molar Mass of KF)

Step1: Identify atomic masses

Atomic mass of K (Potassium) is approximately 39.10 g/mol, and atomic mass of F (Fluorine) is approximately 19.00 g/mol.

Step2: Calculate molar mass of KF

Molar mass of KF = atomic mass of K + atomic mass of F = \( 39.10 + 19.00 = 58.10 \) g/mol.

Part A (Molar Mass of \( Ag_3N \))

Step1: Identify atomic masses

Atomic mass of Ag (Silver) is approximately 107.87 g/mol, and atomic mass of N (Nitrogen) is approximately 14.01 g/mol.

Step2: Calculate molar mass of \( Ag_3N \)

Molar mass of \( Ag_3N \) = (3 × atomic mass of Ag) + atomic mass of N = \( 3\times107.87 + 14.01 = 323.61 + 14.01 = 337.62 \) g/mol.

Question 13 (Moles of \( CaCl_2 \))

Step1: Convert volume to liters

Volume of HCl = 3.65 mL = \( \frac{3.65}{1000} = 0.00365 \) L.

Step2: Calculate moles of HCl

Moles of HCl = Molarity × Volume = \( 0.62 \, M \times 0.00365 \, L = 0.002263 \) mol.

Step3: Use stoichiometry

From the reaction \( 2HCl + CaCO_3
ightarrow CaCl_2 + H_2O + CO_2 \), the mole ratio of \( HCl \) to \( CaCl_2 \) is 2:1.
So, moles of \( CaCl_2 \) = \( \frac{1}{2} \times \) moles of HCl = \( \frac{0.002263}{2} = 0.0011315 \) mol (approx 0.00113 mol).

Answer:

s:

  • Molar mass of KF: \( \boldsymbol{58.10} \) g/mol
  • Molar mass of \( Ag_3N \): \( \boldsymbol{337.62} \) g/mol
  • Moles of \( CaCl_2 \): \( \boldsymbol{0.00113} \) mol (or more precisely 0.0011315 mol)