Answer to Question #75163 in General Chemistry for Chloe Lozano

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

Solution:
3NaSO4 (aq) + 2Al(NO3)3 (aq) → Al2(SO4)3 (s) + 6 NaNO3 (aq)
We have 960.0 ml we can calculate the moles of aluminum nitrate.
5.20 M = x mol / 0.960 L
x mol = 5.20 M x 0.960 L
x mol = 4.99 mol of aluminum nitrate
For every 2 moles of aluminum nitrate that react we get 1 mole of aluminum sulfate (the coefficients
of the reaction) Therefore:
4.99 mol of Aluminum nitrate x 1mol of aluminum sulfate /2 mol aluminum nitrate =2.5 mol
aluminum sulfate
Now convert to grams = 2.5 mol aluminum sulfate x 342.15 g/mol = 855.95 g
Answer: 855.95 g (E. None of the Above)