Chloe S. answered • 09/21/20

Patient & Bright Tutor for Math/Science/English/Spanish/SAT Prep

__Question 1__

For this, we simply need to multiply by a molar ratio. For ever 3 moles of O_{2}, we have 2 moles of KClO_{3}, so:

12.00 mol KClO_{3} * (3 mol O_{2} / 2 mol KClO_{3}) = **18 mol O**_{2}

__Question 2__

We are again performing molar ratios using the coefficients.

a) for every 2 mol C_{4}H_{10}, there are 13 mol O_{2}. So the molar ratio is 2 mol C_{4}H_{10} / 13 mol O_{2}, or 2 : 13.

b) for every 13 mol O_{2}, there are 8 mol CO_{2}. So the molar ratio is 13 mol O_{2} / 8 mol CO_{2}, or 13 : 8.

c) for every 13 mol O2, there are 10 mol H_{2}O. So the molar ratio is 13 mol O2 / 10 mol H_{2}O, or 13 : 10.

d) for every 2 mol C_{4}H_{10}, there are 8 mol CO_{2}. We can simplify by dividing each by 2, so the molar ratio is 1 mol C_{4}H_{10} / 4 mol CO_{2}, or 1 : 4.

e) for ever 2 mol C_{4}H_{10}, there are 10 mol H_{2}O. We can simplify by dividing each by 2, so the molar ratio is 1 mol C_{4}H_{10} / 5 mol H_{2}O, 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 Cl_{2} = 70.90g.

a) the limiting reagent is K because we are told there is excess Cl_{2}. 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 Cl_{2} because we are told there is excess K.

Now we can do our stoichiometric conversion:

1.00g Cl_{2} * (1 mol Cl_{2 }/ 70.90g Cl_{2}) * (2 mol KCl / 1 mol Cl_{2}) * (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 Na_{2}O = 61.98g Na_{2}O. Molar mass NaOH = 40.00g NaOH.

a) 1.20*10^{2}g Na_{2}O * (1mol Na_{2}O / 61.98g Na_{2}O) * (2 mol NaOH / 1 mol Na_{2}O) * (40.00g NaOH / 1 mol NaOH) = 154.89g NaOH

b) 1.60*10^{2}g NaOH * (1 mol NaOH / 40.00g NaOH) * (1 mol Na_{2}O / 2 mol NaOH) * (61.98g Na2O / 1 mol Na_{2}O) = 123.96g Na_{2}O