For each of the following reactions, calculate the grams of indicated product when 23.5 g of the first reactant and 42.0 g of the second reactant is...

For
all of these parts, the same procedure will help you attain the correct answer. I will solve the
first part, and if you follow the procedure, you should be able to solve the remaining
parts.

2SO‚‚ (g) + O‚‚ (g) †’ 2SO‚ƒ (g)

Given: 23.5 g of first reactant (SO‚‚) and 42.0 g of second
reactant (oxygen).

The first step is to determine the moles of each reactant,
as per the given mass.

The molecular mass of SO‚‚ is 64 g/mole, and that of
O‚‚ is 32 g/mole.

The moles of SO‚‚ in 23.5 g are 23.5/64 = 0.37
moles.

Similarly, the moles of O‚‚ in 42 g are 42/32 = 1.31 moles.

Step two is to check the stoichiometric equivalency from the given chemical
reaction.

Here, 2 moles of SO reacts with 1 mole of O‚‚ to generate 2 moles
of SO‚ƒ.

Step 3 is to check if we have adequate moles of each
reactant.

Here, we have 0.37 moles of SO‚‚. Since 2 moles of SO‚‚ react with
1 mole of O‚‚, each mole of SO‚‚ will need only 1/2 or 0.5 moles of O‚‚.

And,
in the given case, 0.37 moles of SO‚‚ will react with only 0.37/2 = 0.185 moles of
O‚‚.

Step 4 is to determine the product quantity as per stoichiometry and the
limiting reactant.

In this case, SO‚‚ is the limiting reactant and will
govern how much SO‚ƒ will be produced.

For 2 moles of SO‚‚, 2 moles of SO‚ƒ
are produced. This means that for each mole of SO‚‚, 1 mole of SO‚ƒ will be produced. For 0.37
moles of SO‚‚, 0.37 moles of SO‚ƒ will be produced.

The final step is to
calculate the grams of the product using its molecular mass.

The molecular
mass of SO‚ƒ is 80 g/moles. Since only 0.37 moles are generated, the produced mass is 0.37*80 =
29.6 g SO‚ƒ.

Using this same method, you can
solve the rest of the problem.

Hope this helps.