Question

Calculate the amount (in grams) of 85% phosphoric acid you need to use for 0.0700 mol of cyclohexanol. Recall that you need 0.4 equivalents of phosphoric acid.

Answer 1

Number of moles of cyclohexanol being used = 0.0700 mol.

Hence, 1 equivalent of phosphoric acid must equal 0.0700 mol of phosphoric acid.

Since we only need 0.4 equivalent of phosphoric acid, the number of moles of phosphoric acid needed is

$\frac{0.0700 \ mol }{1 \ equiv.} \times 0.4 \ equiv. = 0.0280 \ mol$

Molar mass of phosphoric acid ,  $H_3PO_4$ = 97.994 g/mol

Hence, amount of phosphoric acid needed in grams is

$mass = moles \times molar \ mass = 0.0280 \ mol \times 97.994 \frac{g}{mol} \approx 2.744 \ g$

Note that the phosphoric acid available is 85% w/w.

Hence, 100 g of the available phosphoric acid contains only 85.0 g $H_3PO_4$.

Hence, the amount of 85% phosphoric acid that will contains 2.744 g of $H_3PO_4$ is

$\frac{100 \ g}{85.0 \ g} \times 2.744 \ g \approx 3.228 \ g \approx 3.23 \ g$

Hence, we need to use 3.23 g (rounded to three significant figures) of the 85% phosphoric acid.