How many grams of calcium nitrate (molar mass = 164 g) would be formed if 0.500 g of calcium chloride (molar mass = 111 g) were reacted with an excess of silver nitrate (molar mass = 170. g

2AgNO3 + CaCl2 ==> Ca(NO3)2 + 2AgCl

mols CaCl2 = grams/molar mass = ?
Use the coefficients in the balanced equation to convert mols CaCl2 to mols Ca(NO3)2 (That should be mols CaCl2 x 2 = ?).
Then g Ca(NO3)2 = mols Ca(NO3)2 x molar mass Ca(NO3)2 = ?

To determine the number of grams of calcium nitrate formed, we need to first determine the balanced chemical equation for the reaction between calcium chloride and silver nitrate.

The balanced chemical equation for the reaction is:

CaCl2 + 2AgNO3 → Ca(NO3)2 + 2AgCl

From the equation, we can see that one mole of calcium chloride (CaCl2) reacts with one mole of calcium nitrate (Ca(NO3)2). Therefore, we can set up a mole-to-mole ratio to find the number of moles of calcium nitrate formed.

First, calculate the number of moles of calcium chloride using its molar mass:

moles of calcium chloride = mass of calcium chloride / molar mass of calcium chloride
moles of calcium chloride = 0.500 g / 111 g/mol
moles of calcium chloride ≈ 0.00450 mol

Since calcium nitrate and calcium chloride have a 1:1 mole ratio, the number of moles of calcium nitrate formed will also be approximately 0.00450 mol.

Finally, calculate the mass of calcium nitrate formed using its molar mass:

mass of calcium nitrate = moles of calcium nitrate × molar mass of calcium nitrate
mass of calcium nitrate = 0.00450 mol × 164 g/mol
mass of calcium nitrate ≈ 0.738 g

Therefore, approximately 0.738 grams of calcium nitrate would be formed if 0.500 grams of calcium chloride were reacted with an excess of silver nitrate.