1/ determine the number of moles of aluminum permanganate in a 975. gram sample

Question

1/ determine the number of moles of aluminum permanganate in a 975. gram sample of aluminum permanganate.
2/ determine the total number of atoms in a 355. gram sample of calcium acetate.
3/ write a chemical equation for the following statement. solid copper reacts with oxygen gas solid copper(I) oxide forms.
4/4AL(s) + 3O2(g) -----> 2Al2O3(s)
how many grams of aluminum oxide will form when 250. grams aluminum react completely with an excess amount of oxygen gas?
How many grams of oxygen gas will react if 7.50 grams of aluminum react completely
5/ determine the formula mass of silver chromate and the percent by mass of silver in silver chromate.
6/ use the following data to determine the empirical formula and molecular formula of a compound with the following percent composition. 64.8% C , 13.6%H, and 21.6% O. Molar mass= 74g/mol
7/sodium azide decomposes to sodium and nitrogen gas according to the following equation. 2NaN3(s)+3 N2(g)
how many grams of sodium azide are required to produce 15.0 moles of nitrogen gas?
In one experiment, 300. grams of sodium azide produces 150. grams of nitrogen gas. determine the percent yield of this reaction.
8/ a solution prepared by combining 347 grams of calcium hydroxide with enough water to make 2.50 liters of solution. Determine the molarity of the solution.

Explanation / Answer

1) aluminium permanganate = Al(MnO4)3

Molar mass of aluminium permanganate = 383.79 g/mole

Thus, moles of aluminium permanganate in 975 g sample of it = mass/molar mass = 975/383.79 = 2.54

2) Calcium acetate = Ca(C2H3O2)2

Molar mass of Calcium acetate = 158.17 g/mole

Now, moles of Calcium acetate in 355 g of it = mass/molar mass = 355/158.17 = 2.24

Now, 1 atom of Calcium acetate contains 15 atoms

Thus, 1 mole of Calcium acetate conatins 15*6.022*1023 atoms

Hence, 2.24 moles of Calcium acetate contains 2.24*15*6.022*1023 = 2.027*1025 atoms

3) 4Cu(s) + O2(g) -------------> 2Cu2O

4) 4Al(s) + 3O2(g) ------------> 2Al2O3(s)

Molar mass of Al = 27 g/mole

Thus, moles of Al in 250 g of it = mass/molar mass = 250/27 = 9.26

Now, as per the balanced reaction, moles of Al2O3 formed = (1/2)*moles of Al reacted = 4.63

Now, molar mass of Al2O3 = 102 g/mole

Hence mass of Al2O3 formed = moles*molar mass = 102*4.63 = 472.22 g

Now, moles of Al in 7.5 g of it = mass/molar mass = 7.5/27 = 0.278

Now, As per the balanced reaction, moles of oxygen required = (3/4)*moles of Al reacting = 0.208

Now, molar mass of oxygen = 32 g/mole

Hence mass of oxygen reacting with 27 g of Al = 32*0.208 = 6.667 g

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