Question

$H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$

When 46g of I2and 1g of H2are heated at equilibrium at ${450}^{\circ }C$, the equilibrium mixture contained 1.9g of I2.

How many moles of $I_2$ and HI are present at equilibrium.

• A. 0.0075 & 0.147 moles
• B. 0.0050 & 0.147 moles
• C. 0.0075 & 0.347 moles
• D. 0.0052 & 0.347 moles

Answer 1

The correct option is C

0.0075 & 0.347 moles

moles of $I_2$ taken $=\frac{46}{254}=0.181$

moles of $H_2$ taken $=\frac{1}{2}=0.5$

moles of $I_2$ remaining $=\frac{1.9}{254}=0.0075$

moles of $I_2$ used $=0.181-0.0075=0.1735$

moles of $H_2$ used $=0.1735$ moles of $H_2$

remaining $0.5-0.1735=0.3565$

moles of HI formed $=0.1735\times 2=0.347$

At equlibrium

moles of $I_2=0.0075$ moles

moles of HI = 0.347 moles