Help me to answer some stoichiometry question 🙏

Question

1. Given the following equation:

2 KClO3 ---> 2 KCl + 3 O2

How many moles of O2 can be produced by letting 12.00 moles of KClO3 react?

2. Given the following equation:

2 C4H10 + 13 O2 ---> 8 CO2 + 10 H2O, show what the following molar ratios should be.

a. C4H10 / O2

b. O2 / CO2

c. O2 / H2O

d. C4H10 / CO2

e. C4H10 / H2O

3. Given the following equation:

2 K + Cl2 ---> 2 KCl

How many grams of KCl is produced from 2.50 g of K and excess Cl2. From 1.00 g of Cl2 and excess K?

4. Given the following equation:

Na2O + H2O ---> 2 NaOH

How many grams of NaOH is produced from 1.20 x 10^2

grams of Na2O? How many grams of Na2O are required to produce 1.60 x 10^2 grams of NaOH?

Answer 1

Question 1

For this, we simply need to multiply by a molar ratio. For ever 3 moles of O₂, we have 2 moles of KClO₃, so:

12.00 mol KClO₃ * (3 mol O₂ / 2 mol KClO₃) = 18 mol O₂

Question 2

We are again performing molar ratios using the coefficients.

a) for every 2 mol C₄H₁₀, there are 13 mol O₂. So the molar ratio is 2 mol C₄H₁₀ / 13 mol O₂, or 2 : 13.

b) for every 13 mol O₂, there are 8 mol CO₂. So the molar ratio is 13 mol O₂ / 8 mol CO₂, or 13 : 8.

c) for every 13 mol O2, there are 10 mol H₂O. So the molar ratio is 13 mol O2 / 10 mol H₂O, or 13 : 10.

d) for every 2 mol C₄H₁₀, there are 8 mol CO₂. We can simplify by dividing each by 2, so the molar ratio is 1 mol C₄H₁₀ / 4 mol CO₂, or 1 : 4.

e) for ever 2 mol C₄H₁₀, there are 10 mol H₂O. We can simplify by dividing each by 2, so the molar ratio is 1 mol C₄H₁₀ / 5 mol H₂O, or 1 : 5.

Question 3

First lay out our molar masses. Molar mass of K = 39.10g. Molar mass of KCl = 74.55g. Molar mass of Cl₂ = 70.90g.

a) the limiting reagent is K because we are told there is excess Cl₂. Now we can do our stoichiometric conversion:

2.5g K * (1mol K / 39.10g K) * (2 mol KCl / 2 mol K) * (74.55g KCl / 1 mol KCl) = 4.77g KCl

b) the limiting reagent is Cl₂ because we are told there is excess K.

Now we can do our stoichiometric conversion:

1.00g Cl₂ * (1 mol Cl₂ / 70.90g Cl₂) * (2 mol KCl / 1 mol Cl₂) * (74.55g KCl / 1 mol KCl) = 2.10g KCl

Question 4

We again lay out our molar masses. We know that water is going to be used in excess. Molar mass of Na₂O = 61.98g Na₂O. Molar mass NaOH = 40.00g NaOH.

a) 1.20*10²g Na₂O * (1mol Na₂O / 61.98g Na₂O) * (2 mol NaOH / 1 mol Na₂O) * (40.00g NaOH / 1 mol NaOH) = 154.89g NaOH

b) 1.60*10²g NaOH * (1 mol NaOH / 40.00g NaOH) * (1 mol Na₂O / 2 mol NaOH) * (61.98g Na2O / 1 mol Na₂O) = 123.96g Na₂O