Question
The industrial production of hydriodic acid takes place bytreatment of iodine with hydrazine $\left(\mathrm{N}_{2} \mathrm{H}_{4}\right) :$$2 \mathrm{I}_{2}+\mathrm{N}_{2} \mathrm{H}_{4} \longrightarrow 4 \mathrm{HI}+\mathrm{N}_{2}$(a) How many grams of $\mathrm{I}_{2}$ are needed to react with 36.7 $\mathrm{g}$ of $\mathrm{N}_{2} \mathrm{H}_{4} ?$(b) How many grams of HI are produced from the reaction of 115.7 $\mathrm{g}$ of $\mathrm{N}_{2} \mathrm{H}_{4}$ with excess iodine?
Step 1
We do this by dividing the given mass by the molar mass of $\mathrm{N}_{2} \mathrm{H}_{4}$, which is 32.5 g/mol. So, we have $\frac{36.7 \, \text{g}}{32.5 \, \text{g/mol}} = 1.13 \, \text{mol}$ of $\mathrm{N}_{2} \mathrm{H}_{4}$. Show more…
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The industrial production of hydriodic acid takes place by treatment of iodine with hydrazine $\left(\mathrm{N}_{2} \mathrm{H}_{4}\right)$ : $$ 2 \mathrm{I}_{2}+\mathrm{N}_{2} \mathrm{H}_{4} \longrightarrow 4 \mathrm{HI}+\mathrm{N}_{2} $$ (a) How many grams of $\mathrm{I}_{2}$ are needed to react with $36.7 \mathrm{~g}$ of $\left(\mathrm{N}_{2} \mathrm{H}_{4}\right) ?$ (b) How many grams of HI are produced from the reaction of $115.7 \mathrm{~g}$ of $\mathrm{N}_{2} \mathrm{H}_{4}$ with excess iodine?
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The industrial production of hydriodic acid takes place bytreatment of iodine with hydrazine $\left(\mathrm{N}_{2} \mathrm{H}_{4}\right) :$ $2 \mathrm{I}_{2}+\mathrm{N}_{2} \mathrm{H}_{4} \longrightarrow 4 \mathrm{HI}+\mathrm{N}_{2}$ (a) How many grams of $\mathrm{I}_{2}$ are needed to react with 36.7 $\mathrm{g}$ of $\mathrm{N}_{2} \mathrm{H}_{4} ?$ (b) How many grams of HI are produced from the reaction of 115.7 $\mathrm{g}$ of $\mathrm{N}_{2} \mathrm{H}_{4}$ with excess iodine?
The industrial production of hydriodic acid takes place by treatment of iodine with hydrazine (N2H4): 2I2 + N2H4 -> 4HI + N2 a. How many grams of I2 are needed to react with 34.9 g of (N2H4)? b. How many grams of HI are produced from the reaction of 160.7 g of N2H4 with excess iodine?
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