Chem help please!?

Question

Final tomorrow and still can't figure it out.

3 H2SO4 + 2 Al ------> Al2(SO4)3 + 3H2

Given 700. grams of HSO4. Calculate

1. moles of H2SO4

2. moles of Al reacted

3. grams of Al reacted

4. moles of Al2(SO4)3 formed

5. grams of Al2(SO4)3 formed

6. moles of H2 formed

7. grams of H2 formed

Answer 1

1) number of moles = 700/98 = 7.14
98 is molecular weight H2SO4

2) for 3 moles of the acid you have 2 of Al

so number of moles of Al = 7.14*2/3 = 4.76

3) the weight of Al is 27 so 4.76 *27 = 128.6 g

4) you see 2Al will give 1 Al2(SO4)3 from the equation

hence 4.76 moles give 2.38 moles of Al2(SO4)3

5) mass of the salt is 2*27 +(32+4*16)*3 =342g that you multiply by 2.38 = 914 g

6)Looking the equation you see that the coefficients
of H2SO4 and H2 are the same 3 So number ofmoles are the same 7.14


7) a mole of H2 = 2g so 7.14 moles = 14.18 g

Answer 2

1) number of moles = 700/98 = 7.14
98 is molecular weight H2SO4

2) for 3 moles of the acid you have 2 of Al

so number of moles of Al = 7.14*2/3 = 4.76

3) the weight of Al is 27 so 4.76 *27 = 128.6 g

4) you see 2Al will give 1 Al2(SO4)3 from the equation

hence 4.76 moles give 2.38 moles of Al2(SO4)3

5) mass of the salt is 2*27 +(32+4*16)*3 =342g that you multiply by 2.38 = 914 g

6)Looking the equation you see that the coefficients
of H2SO4 and H2 are the same 3 So number of moles are the same 7.14


7) a mole of H2 = 2g so 7.14 moles = 14.18
i hope that helps ur answer

Answer 3

1. To find the moles of H2SO4: set up a conversion factor with 700. g H2SO4 * 1 mole H2SO4 / 98.09 g H2SO4 (the molar mass of sulfuric acid) and that should equal 7.14 moles of H2SO4. (This step is dimensional analysis)

2. To find the moles of Al: use your BALANCED chemical equation to find that 3 moles of H2SO4 react with 2 moles of Al, then set up another conversion factor that looks like; 7.14 mol H2SO4 * 2 mol Al / 3 mol H2SO4 and that should equal 4.76 moles of Al. (This step is stoichiometry)

3. To find the grams of Al: set up a conversion factor with the grams of Al in one mole multiplied by the moles of Al. 4.76 mol Al * 27.00 g Al / 1 mol Al = 129 g Al (And back to dimensional analysis)

4-7 are done exactly the same way as 2 and 3, but not with the same numbers.

4. 7.14 mol H2SO4 * 1 mol Al2(SO4)3 / 3 mol H2SO4 = 2.38 mol Al2(SO4)3

5. 2.38 mol Al2(SO4)3 * 342.2 g Al2(SO4)3 / 1 mol Al2(SO4)3 = 814 g Al2(SO4)3

6. 7.14 mol H2SO4 * 3 mol H2 / 3 mol H2SO4 = 7.14 mol H2

7. 7.14 mol H2 * 2.016 g H2 / 1 mol H2 = 14.4 g H2