Answer to Question #165972 in General Chemistry for Ella Bennett

Question #165972

question 1: 4Al + 3O₂ → 2Al₂O₃ How many grams of Aluminum are needed if you have 1.20 moles of aluminum oxide? *10 points43.2 g64.8 g57.6 g173 g
question 2: 4Fe + 3O₂ → 2Fe₂O₃ Given: 5.78 moles of iron, how many moles of oxygen are needed? *10 points4.34 mol3.78 mol2.84 mol90.7 mol
question 3: N₂(g) + 3H₂(g) 2NH₃(g) There is 6.41 grams of hydrogen, how many moles of ammonia are present? *10 points2.27 mol1.14 mol2.14 mol2.56 mol
question 4: 2P₂O₅ + 6H₂O 4H₃PO₄ Given: 4.5 grams of H₂O, how many moles of H₃PO₄ are needed? *10 points0.083 mol0.75 mol1.5 mol0.17 mol
question 5: 2P₂O₅ + 6H₂O 4H₃PO₄ If you have 1.5 moles of H₂O, how many grams of P₂O₅ would you need? *10 points1917 g213 g71 g98 g
question 6: F₂ + 2 NaBr → 2 NaF + Br₂ What is the correct setup of ratios to determine the mass of Bromine produced when 0.789 moles of sodium bromide are present? *10 points0.789 x (83/1) x (2/1)0.789 x (1/83) x (1/2)0.789 x (1/2) x (160/1)0.789 x (2/1) x (160/1)
question 7: 2Al + 3Cl₂ 2AlCl₃ If you have 6.5 mol of Al, how many grams of Cl₂ are needed? *10 points480 g213 g240 g692 g
question 8; 4Al + 3O₂ → 2Al₂O₃ Given 6.20 moles of aluminum oxide, how many grams of oxygen are needed? *10 points111 g250 g298 g335 g
question 9: F₂ + 2 NaBr → 2 NaF + Br₂ What is the correct setup of ratios to determine the mass of sodium fluoride produced when 0.789 moles of fluorine are present? *10 points0.789 x (2/2) x (42/1)0.789 x (2/1) x (1/42)0.789 x (1/2) x (38/1)0.789 x (2/1) x (42/1)
question 10: N₂(g) + 3H₂(g) 2NH₃(g) There is 5.41 grams NH₃, how many moles of hydrogen are needed? *10 points454 mol0.477 mol0.580 mol16.2 mol



please answer all of these, please



1
Expert's answer
2021-02-28T06:20:16-0500

1. 108 x 1.2/2= 64.8 g.

2. 5.78 x 3/4= 4.335 moles.

3. 6.41 x 2/6=2.133333 moles.

4. 4.5 x 4/6x18= 0.167 moles.

5. 1.5 x 4x98/6 = 98 g.

6. 0.789/2 = 0.3945 moles.((1/2)0.789)

7. 6.5 x 3 x 71/2= 692.5 g.

8. 3 x 32 x 6.2/4 = 148.8 g.

9. 38 x 0.789 /2 = 14.991 g. ((1/2) x (38/1)0.789).

10. 5.41 x 3/34=0.477 moles.


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