Question #7a6ab

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Question #7a6ab

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Answer 1 · 2015-03-14T00:10:51

$33.3 g HI$ $"33.3 g HI"$ will be produced.

Start with a balanced equation:

$H_{2} (g)$ $"H"_2("g")$ + $I_{2} (s)$ $"I"_2(s)$ $\to$ $rarr$ $2HI(g)$ $"2HI(g)"$

Convert 33.0 grams of iodine to moles of iodine by dividing the given mass by its molar mass. The molar mass of $I_{2}$ $"I"_2$ is 253.80894 g/mol.

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

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

Determine the number of moles of $HI$ $"HI"$ that can be produced from ${0.13001 mol I}_{2}$ $"0.13001 mol I"_2$ by multiplying times the mole ratio so that $HI$ $"HI"$ is on top.

${0.13001 mol I}_{2}$ $"0.13001 mol I"_2$ x $\frac{2 mol HI}{1 mol I2}$ $"2 mol HI"/"1 mol I2"$ = $0.26002 mol HI$ $"0.26002 mol HI"$

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

$0.26002 mol HI$ $"0.26002 mol HI"$ x $\frac{127.91247 g HI}{1 mol HI}$ $"127.91247 g HI"/"1 mol HI"$ = $33.259 g HI$ $"33.259 g HI"$ = $33.3g HI$ $"33.3g HI"$ due to 3 sig figs in 33.0 g of $I_{2}$ $"I"_2"$ stated in the problem.

Answer 2 · 2015-03-14T00:20:05

33.26 g of $H I$ $HI$ will be formed.

Explanation:

33.26 g of HI will be formed.

$H_{2 (g)} + I_{2 (g)} \to 2 H I_{(g)}$ $H_(2(g))+I_(2(g))rarr2HI_((g))$

1mol + 1mol $\to$ $rarr$ 2mol

Using approximate $A_{r}$ $A_r$ values:

$(2 \times 127) g I_{2} \to 2 (1 + 127) g H I$ $(2xx127)gI_2rarr2(1+127)gHI$

So:

$254 g \to 256 g$ $254grarr256g$

So:

$1 g \to \frac{256}{254} g$ $1grarr256/254g$

So:

$33.0 g \to \frac{256}{254} \times 33.0 = 33.26 g$ $33.0grarr(256)/(254)xx33.0= 33.26g$

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