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Lubov Fominskaja [6]
3 years ago
10

Aluminum reacts with sulfuric acid to produce aluminum sulfate and hydrogen gas. how many grams of aluminum sulfate would be for

med if 250 g hso completely reacted with aluminum?
Chemistry
2 answers:
faust18 [17]3 years ago
4 0
Atomic weights S-32 O-16 Al-27 H-1
aluminium sulfate 250g= 2.55 mol  (250/98) 
3Al +  2H2SO4 --------->  Al2(SO4)3 + 2H2 
therefore  two parts of sulfuric acid produces one part of aluminum sulfate. 

2.55mol /2 =1.275 mol  1.275mols of aluminium sulfate is produced.  

1.275mol s multiplied by aluminium sulfate's molar mass (342g/mol) gives it's mass in grams.. 

how many grams of aluminium sulfate produced ? 436.22 g  HOPE THIS HELPS GOOD LUCK  
sesenic [268]3 years ago
4 0

Answer:

m_{Al_2(SO_4)_3}=291gAl_2(SO_4)_3

Explanation:

Hello,

In this case, the undergoing reaction is:

2Al(s)+3H_2SO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+3H_2(g)

In such a way, from the 250 g of sulfuric acid that completely reacts with the aliminium, the yielded grams of aluminium sulfate are computed by considering the 3 to 1 molar relationship between sulfuric acid and aluminium sulfate and the corresponding molar masses of 98 g/mol and 342 g/mol respectively:

m_{Al_2(SO_4)_3}=250gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}*\frac{1molAl_2(SO_4)_3}{3molH_2SO_4}  *\frac{342gAl_2(SO_4)_3}{1molAl_2(SO_4)_3} \\\\m_{Al_2(SO_4)_3}=291gAl_2(SO_4)_3

Best regards.

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8 0
4 years ago
If the Ksp of NaCl is experimentally determined to be 53.9, then what is the concentration of Na (in M) when it begins to crysta
Lady bird [3.3K]

For the concentration of Na in M when it begins to crystallize out of the solution is mathematically given as

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<h3>what is the concentration of Na in M when it begins to crystallize out of the solution?</h3>

Generally, the equation for the Ksp of the solution is mathematically given as

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In conclusion, the concentration of Na in M when it begins to crystallize out of the solution

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brainly.com/question/7185695

4 0
3 years ago
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Answer:

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Explanation:

given:

density of a substance is 3.4 x 10^3 g/mL

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find:

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-------------------------------------------------

let density (a) = 3.4 x 10^3 g/mL

density(b) = 1.5 x 10^1 g/mL

since there is no specific density provided for the mixture, we then add both to increase the density,

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total density = 3.4 x 10^3 g/mL + 1.5 x 10^1 g/mL

total density = 3.415 x 10^3 g/mL

therefore,

the total density of the mixture = 3.415 x 10^3 g/mL

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