Using the balanced equation below, how many grams of cesium fluoride would be required to make 73.1 g of cesium xenon heptafluoride?

CsF+XeF6 ---> CsXeF7

It's 28

To determine the number of grams of cesium fluoride (CsF) required to make 73.1 g of cesium xenon heptafluoride (CsXeF7), you need to use the balanced equation and the molar mass of each compound.

1. Begin by finding the molar masses of CsF and CsXeF7:
- Molar mass of CsF:
- Molar mass of Cesium (Cs) = 132.90545 g/mol
- Molar mass of Fluorine (F) = 18.9984032 g/mol
- Molar mass of CsF = (molar mass of Cs) + (molar mass of F) = 132.90545 g/mol + 18.9984032 g/mol = 151.9038532 g/mol

- Molar mass of CsXeF7:
- Molar mass of Cesium (Cs) = 132.90545 g/mol
- Molar mass of Xenon (Xe) = 131.293 g/mol
- Molar mass of Fluorine (F) = 18.9984032 g/mol
- Molar mass of CsXeF7 = (molar mass of Cs) + (molar mass of Xe) + 7 * (molar mass of F) = 132.90545 g/mol + 131.293 g/mol + 7 * 18.9984032 g/mol = 672.1748756 g/mol

2. Use stoichiometry to calculate the amount of CsF required:
- Balanced equation: CsF + XeF6 -> CsXeF7
- The coefficients in the balanced equation represent the ratio of moles.
- From the balanced equation, the ratio of CsF to CsXeF7 is 1:1.
- To find the moles of CsXeF7: moles = mass / molar mass
- Moles of CsXeF7 = 73.1 g / 672.1748756 g/mol = 0.1088 mol
- Since the stoichiometric ratio is 1:1, the moles of CsF required will also be 0.1088 mol.

3. Convert moles of CsF to grams:
- Grams of CsF = moles * molar mass
- Grams of CsF = 0.1088 mol * 151.9038532 g/mol = 16.5 g

Therefore, you would need 16.5 grams of cesium fluoride to make 73.1 grams of cesium xenon heptafluoride.

1. Convert 73.1 g CsXeF7 to mols. mols = grams/molar mass = ?

2. Using the coefficients in the balanced equation, convert mols CsXeF7 to mols CsF.

3. Now convert mols CsF to grams. g = mols x molar mass.

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