Chem 12.2- Chemical Calculations

Mole ratios from balanced equations may be used to solve problems with other units such as numbers of ______ and _____ of gases at STP

Representaive particles; volumes

The _________ from the balanced equation are used to write conversion factors called ___________.

Coefficients; mole ratios

These conversion factors are used to calculate the number of moles of _________ from a given number of moles of __________.

Product; reactant

In mass-mass calculations, the molar mass is used to convert mass to ______.

Moles

In mass-mass calculations, the molar mad is used to convert mass to moles.

NT, AT, ST

AT

The mole ratio 2 mol HF/1 mol SnF₂ can be used to determine the mass of SnF₂ produced according to the equation:

Sn(s) + 2HF(g) ——-> SnF₂ + H₂(g)

NT, AT, ST

NT; you can't use that ratio because the equation is not balanced

In a volume-volume problem, the 22.4 L/mol factors always cancel out.

NT, AT, ST

AT

In stoichiometric problems, volume is expressed in terms of liters.

NT, AT, ST

ST; volume could also be expressed in moles or cubic meters

For a mass-mole problem, the first conversion from mass to moles is skipped.

NT, AT, ST

NT; you must convert mass to moles for stoichiometry to work in mass-mole problems (you can skip the step in volume-mole problems)

For a mass-mass problem, the first conversion is from moles to mass.

NT, AT, ST

NT; it is from mass to moles

Because mole ratios from balanced equations are exact numbers, they do not enter into the determination of significant figures.

NT, AT, ST

AT

Moles 0₂ —> grams O₂

Mol * 32.0g/ mol

Liters SO₂ —> grams SO₂ at STP

Liters mol/22.4L 64.1g/mol

Molecules He —> liters He(g) at STP

Molecules mol/(6.0210^23) * 22.4L/mol

Grams Sn —> molecules Sn

Grams mol/119g (6.02*10^23 molecules)/ mol

Molecules H₂O —> grams H₂O

Molecules mol/ (6.0210^23 molecules) * 18.0g/mol

How many liters of carbon monoxide at STP are needed to react with 4.8 g of oxygen gas to produce carbon dioxide?

2CO(g) + O₂(g) —> 2CO₂(g)

6.7 liters of CO

What mass of ammonia, NH3, is necessary to react with 2.1*10^24 molecules of oxygen in the following reaction?

4NH₃(g)+7O₂(g) —> 6H₂O(g) + 4NO₂(g)

33.9 grams of NH3