This question is in my General Chem 2 class. I will give Brainlist!


Given: 8As + 12H20 —>3H2AsO4- + 3H+ + 5AsH3 What mass of arsenic is required to produce 15.0 mL of 0.0852M H2AsO4-?

A. 0.0359 g

B. 0.255 g

C. 17.0 g

D.1134 g

Respuesta :

Answer:

B

Explanation:

From the given chemical equation, we want to determine the mass of arsenic necessary to produce 15.0 mL of 0.0852 M H₂AsO₄⁻.

Find the amount of moles present in 15.0 mL of 0.0852 M H₂AsO₄⁻ solution. Recall that M = mol/L:

[tex]\displaystyle 15.0\text{ mL} \cdot \frac{0.0852\text{ mol H$_2$AsO$_4^-$}}{1\text{ L}} \cdot \frac{1\text{ L}}{1000\text{ mL}} = 0.00128\text{ mol H$_2$AsO$_4^-$}[/tex]

With stoichiometric ratios and molecular weights, determine the mass of arsenic necessary:

[tex]\displaystyle 0.00128 \text{ mol H$_2$AsO$_4^-$} \cdot \frac{8\text{ mol As}}{3\text{ mol H$_2$AsO$_4^-$}}\cdot \frac{74.92\text{ g As}}{1\text{ mol As}} = 0.256\text{ g As}[/tex]

In conclusion, the answer is B.