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lana66690 [7]
3 years ago
8

A 100.0 mL sample of 0.200 M HCl is mixed with a 100.0 mL sample of 0.205 M NaOH in a coffee cup calorimeter. If both solutions

were initially at 25.0 C and the temperature of the resulting solution was recorded as 27.0 C, determine the delta Hrxn (in kJ/mol). Assume no heat is lost to the calorimeter. Hint: What is the limiting reagent?
Chemistry
1 answer:
Bas_tet [7]3 years ago
8 0
I just need more points that's why I'm doing this
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How many grams of aluminum sulfate: Al2(SO4)3 , would be formed if 250 g of H2SO4 react with aluminum? The reaction is: 2Al + 3H
NikAS [45]

Answer:

290.8 grams of aluminium sulfate (Al2(SO4)3) will be produced.

Explanation:

Step 1: Data given

Mass of H2SO4 = 250 grams

Molar mass H2SO4 = 98.08 g/mol

Step 2: The balanced equation

2Al + 3H2SO4 → Al2(SO4)3 + 3H2

Step 3: Calculate moles H2SO4

Moles H2SO4 = mass H2SO4 / molar mass H2SO4

Moles H2SO4 = 250 grams / 98.08 g/mol

Moles H2SO5 = 2.55 moles

Step 4: Calculate moles Al2(SO4)3

For 2 moles Al we need 3 moles H2SO4 to produce 1 mol Al2(SO4)3 and 3 moles H2

For 2.55 moles H2SO4 we'll have 2.55/3 = 0.85 moles Al2(SO4)3

Step 5: Calculate mass Al2(SO4)3

Mass Al2(SO4)3 = moles Al2(SO4)3 * molar mass

Mass Al2(SO4)3 = 0.85 moles * 342.15 g/mol

Mass  Al2(SO4)3 = 290.8 grams

290.8 grams of aluminium sulfate (Al2(SO4)3) will be produced.

7 0
4 years ago
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