# Question 7a6ab

Mar 14, 2015

$\text{33.3 g HI}$ will be produced.

"H"_2("g")# + ${\text{I}}_{2} \left(s\right)$ $\rightarrow$ $\text{2HI(g)}$

Convert 33.0 grams of iodine to moles of iodine by dividing the given mass by its molar mass. The molar mass of ${\text{I}}_{2}$ is 253.80894 g/mol.

$\text{33.0 g I"_2}$$\div$$\text{253.80894 g/mol}$ = ${\text{0.13001 mol I}}_{2}$

From the balanced equation, the mole ratio of ${\text{I}}_{2}$ to $\text{HI}$ is ${\text{1 mol I"_2": 2 mol I}}_{2}$.

Determine the number of moles of $\text{HI}$ that can be produced from ${\text{0.13001 mol I}}_{2}$ by multiplying times the mole ratio so that $\text{HI}$ is on top.

${\text{0.13001 mol I}}_{2}$ x $\text{2 mol HI"/"1 mol I2}$ = $\text{0.26002 mol HI}$

Calculate the mass of $\text{HI}$ in grams by multiplying the mol $\text{HI}$ times its molar mass. The molar mass of $\text{HI}$ is $\text{127.91247 g/mol}$.

$\text{0.26002 mol HI}$ x $\text{127.91247 g HI"/"1 mol HI}$ = $\text{33.259 g HI}$ = $\text{33.3g HI}$ due to 3 sig figs in 33.0 g of $\text{I"_2}$ stated in the problem.

Mar 14, 2015

33.26 g of $H I$ will be formed.

#### Explanation:

33.26 g of HI will be formed.

${H}_{2 \left(g\right)} + {I}_{2 \left(g\right)} \rightarrow 2 H {I}_{\left(g\right)}$

1mol + 1mol $\rightarrow$ 2mol

Using approximate ${A}_{r}$ values:

$\left(2 \times 127\right) g {I}_{2} \rightarrow 2 \left(1 + 127\right) g H I$

So:

$254 g \rightarrow 256 g$

So:

$1 g \rightarrow \frac{256}{254} g$

So:

$33.0 g \rightarrow \frac{256}{254} \times 33.0 = 33.26 g$