Question

For the reaction 3NaSO4 (aq) + 2Al(NO3)3 (aq) → Al2(SO4)3 (s) + 6 NaNO3 (aq), adding 960.0 ml of 5.20 M Aluminum Nitrate to excess Sodium Sulfate will produce how many grams of Aluminum Sulfate?

A. 4.99 mol Al(NO3)3
B. 856 mol Al2(SO4)3
C. 296.4 g Al2(SO4)3
D. 499 g Al(NO3)3
E. None of the Above

Answer 1

Question # 75421, answer

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For the reaction 3NaSO4 (aq) + 2Al(NO3)3 (aq) → Al2(SO4)3 (s) + 6 NaNO3 (aq), adding 960.0 ml of 5.20 M Aluminum Nitrate to excess Sodium Sulfate will produce how many grams of Aluminum Sulfate?

A. 4.99 mol Al(NO3)3

B. 856 mol Al2(SO4)3

C. 296.4 g Al2(SO4)3

D. 499 g Al(NO3)3

E. None of the Above

Answer:

Calculate moles of Aluminum Nitrate Al(NO3)3 = 5.20 moles/L x 0.96 L = 4.992 moles

Therefore, moles of Aluminum Sulfate Al2(SO4)3 = 4.992/ 2 = 2.496 moles

Grams of Aluminum Sulfate Al2(SO4)3 = 342.15 g/mol x 2.496 moles = 854 g Al2(SO4)3

Therefore, correct choice is E. None of the Above