D. 72.4 g
Step 1: Calculate the molar mass of AB (A=50 g/mol, B=60 g/mol) and C2 (C=25 g/mol).
AB: 50 g/mol + 60 g/mol = 110 g/mol
C2: 25 g/mol x 2 = 50 g/mol
Step 2: Balance the chemical equation AB + C2 -> AC3
3AB + 2C2 -> 3AC3
Step 3: Calculate the molar mass of AC3 (A=50 g/mol, C=15 g/mol)
AC3: 50 g/mol + 3(15 g/mol) = 95 g/mol
Step 4: Calculate the moles of AB present
17.6 g AB x (1 mol AB / 110 g AB) = 0.16 mol AB
Step 5: Use the mole ratio from the balanced equation to find the moles of AC3 produced
0.16 mol AB x (3 mol AC3 / 3 mol AB) = 0.16 mol AC3
Step 6: Calculate the mass of AC3 produced
0.16 mol AC3 x (95 g AC3 / 1 mol AC3) = 15.2 g AC3
Therefore, the mass in grams of AC3 produced when 17.6 g of AB reacts with unlimited C2 is 15.2 g.
Using the same hypothetical, unbalanced equation from problem 8, what mass in grams of AC3 is produced when 17.6 g of AB reacts with unlimited C2?
A. 36.2 g
B. 54.4 g
C. 61.2 g
1 answer