Question

Al(s) + HCl(aq) → AlCl3(aq) + H2(g) 2. According to the equation above, how many grams of aluminum are needed to completely react with 2.19 mol of hydrochloric acid? 3. Sulfur dioxide will react with water to form sulfurous acid SO2(g) + H2O(l) → H2SO3(l) What mass of sulfur dioxide is needed to prepare 27.86 g of H2SO3(l)? 4. The compound P4S3 is used in matches. It reacts with oxygen to produce P4O10 and SO2. The chemical equation is shown below. P4S3(s) + O2(g) → P4O10(s) + SO2(g) What mass of SO2 is produced from the combustion of 0.401 g P4S3? 5. How many moles of Mg3P2(s) can be produced from the reaction of 0.14 mol Mg(s) with 0.020 mol P4(s)? Mg(s) + P4(s) → Mg3P2(s) 6. A 5.95-g sample of AgNO3 is reacted with excess BaCl2 according to the equation AgNO3(aq) + BaCl2(aq) → AgCl(s) + Ba(NO3)2(aq) to give 4.48 g of AgCl. What is the percent yield of AgCl? 7. Sulfur hexafluoride is produced by reacting elemental sulfur with fluorine gas. S8(s) + F2(g) → SF6(g) What is the percent yield if 18.3 g SF6 is isolated from the reaction of 10.0 g S8 and 30.0 g F

Answer 1

1) Al(s) + HCl(aq) → AlCl3(aq) + H2(g)

According to the equation above, how many grams of aluminum are needed to completely react with 2.19 mol of hydrochloric acid

Solution : As per stoichiometry , 1 mole of Aluminium is required to react completely with 1 mole of HCl

so 2.19 moles of Aluminium will be requried to give 2.19 moles of HCl

Atomic weight of Al = 27 g / mole

so weight of 2.19 moles of Aluminium = 2.19 X 27 = 59.13 grams

3. Sulfur dioxide will react with water to form sulfurous acid

SO2(g) + H2O(l) → H2SO3(l) What mass of sulfur dioxide is needed to prepare 27.86 g of H2SO3(l)

Solution:

As per stoichiometry , 1 mole of SO2 is required to produce 1 mole of H2SO3

Or 64 grams will be required for 82 grams of H2SO3

So for 1 gram of H2SO3 , SO2 required = 64 / 82 grams = 0.78 grams

So for 27.86 grams of H2SO3 , we require = 64 X 27.86 / 82 grams of SO2 = 21.74 grams of SO2

4) P4S3(s) + O2(g) → P4O10(s) + SO2(g) What mass of SO2 is produced from the combustion of 0.401 g P4S3

As per stoichiometry, 1 mole of P4S3 gives 1 mole of SO2

Molecular weight of P4S3 = 220 grams / mole

Molecular weight of SO2 = 64 grams / mole

220 grams of P4S3 will give 64 grams of SO2

1 gram will give = 64 / 220 grams of SO2

So 0.401 grams will give = 64 X 0.401 / 220 grams of SO2 = 0.117 grams of SO2

5) How many moles of Mg3P2(s) can be produced from the reaction of 0.14 mol Mg(s) with 0.020 mol P4(s)?

Mg(s) + P4(s) → Mg3P2(s)

Let us balance the equaiton first

3Mg + 1/2 P4 --> Mg3P2

So

By reaction of 3 moles of Mg with 0.5 moles of P4 we obtain 1 mole of Mg3P2

Ration will be Mg / P4 = 3 / 0.5 = 6:1

Let us find the limiting reagent here

The ration given is = 0.14 / 0.02 = 7:1

So the limiting reagent is P4

So 0.02 moles of P4 will give 0.04 moles of Mg3P2