H2(g) + I2(g) leftharpoons 2HI(g) When 46 g of I2 and 1 g of H2 gas heated at equilibrium at 450°C, the equilibrium mixture contained 1.9 g of I2. How many moles of I2 and HI are present at equilibrium?

Question

H2(g) + I2(g) leftharpoons 2HI(g) When 46 g of I2 and 1 g of H2 gas heated at equilibrium at 450°C, the equilibrium mixture contained 1.9 g of I2. How many moles of I2 and HI are present at equilibrium? (A) 0.0075 and 0.147 moles (B) 0.0050 and 0.147 moles (C) 0.0075 and 0.347 moles (D) 0.0052 and 0.347 moles

Tardigrade Admin · Accepted Answer

Moles of I2, taken =(46/254)= 0.181 Moles of H2 taken = (1/2)= 0.5 Moles of I2 remaining = (1.9/254) =0.0075 Moles of I2 used = 0.181 - 0.0075 = 0.1735 Moles of H2 used = 0.1735 Moles of H2 remaining = 0.5 - 0.1735 = 0.3265 Moles of HI formed = 0.1735 × 2 = 0.347 At equilibrium, moles of I2 = 0.0075 moles Moles of HI = 0.347 moles

Q. $H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$
When $46$ g of $I_{2}$ and $1 g$ of $H_{2}$ gas heated at equilibrium at $450° C,$ the equilibrium mixture contained $1.9 g$ of $I_{2}$ . How many moles of $I_{2}$ and $H I$ are present at equilibrium?

$0.0075$ and $0.147$ moles

$0.0050$ and $0.147$ moles

$0.0075$ and $0.347$ moles

$0.0052$ and $0.347$ moles

Q. H2​(g)+I2​(g)⇌2HI(g) When 46 g of I2​ and 1g of H2​ gas heated at equilibrium at 450°C, the equilibrium mixture contained 1.9g of I2​. How many moles of I2​ and HI are present at equilibrium?

0.0075 and 0.147 moles

0.0050 and 0.147 moles

0.0075 and 0.347 moles

0.0052 and 0.347 moles

Q. $H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$
When $46$ g of $I_{2}$ and $1 g$ of $H_{2}$ gas heated at equilibrium at $450° C,$ the equilibrium mixture contained $1.9 g$ of $I_{2}$ . How many moles of $I_{2}$ and $H I$ are present at equilibrium?

$0.0075$ and $0.147$ moles

$0.0050$ and $0.147$ moles

$0.0075$ and $0.347$ moles

$0.0052$ and $0.347$ moles