The minimum mass of mixture of A2 and B4 required to produce at least 1 kg of each product is : (Given At. mass of 'A' = 10 ; At. mass of 'B' = 120) 5A2 + 2B4 → 2AB2 + 4A2B (A) 2120 gm (B) 1060 gm (C) 560 gm (D) 1660 gm

							
. . . . . . . . . . . . . . . . . . . .1000g. . . . . .1000g 
5A2. . . +. . . 2B4 . . .→. . 2AB2. . +. . 4A2B 
(20g/mol). . (480g/mol). . .(250g/mol). . .(140g/mol) 
. . . . . . . . . . . . . . . . . . . 4mol. . . . . . .7.14mol 
if 4 mol of AB2 is produced then 8 mol of A2B is needed (mole ratio 1 to 2) 
so we need more A2B to assure a minimum of 1kg for AB2 
so AB2 is the limiting product 

200g. . . . . .1920g. . . . . . .1000g. . . . . .1120g 
5A2. . . +. . . 2B4 . . .→. . 2AB2. . +. . 4A2B 
(20g/mol). . (480g/mol). . .(250g/mol). . .(140g/mol) 
10mol . . . . 4mol . . . . . . . 4mol. . . . . . .8 mol 
. .|. . . . . . . . |___1 to 1_____| 
. .|______5 to 2___________.| 

so you need 200+1920 = 2120 g of a mix of A2 and B4 (2.12kg)
						

							
Molar masses of AB2​andA2​B are 250 g/mol and 140 g/mole respectively.
1 kg of AB2​ corresponds to 4 moles. It requires 10 mole A2​  and 4 mole B4​ respectively.
1 kg of A2​B corresponds to 7.14 mole. It requires 8.9 mole A2​  and 3.57 mole B4​ respectively.
Thus 10 mole A2​ (200 g) and 4 mole B4​ (1920  g) will produce atleast 1 kg of each product.
The minimum mass of mixture required is 200g+1920g=2120g