Question

we wish to determine how many grams of $\ce{ag_{2}cro_{4}}$ precipitate when 150. ml of 0.500 m $\ce{agno_{3}}$ solution is added to excess $\ce{k_{2}cro_{4}}$

$\ce{2agno_{3}(aq) + k_{2}cro_{4}(aq) \
ightarrow ag_{2}cro_{4}(s) + 2kno_{3}(aq)}$

in the previous step, you determined 0.0750 mol $\ce{agno_{3}}$ react. the molar mass of $\ce{ag_{2}cro_{4}}$ is 331.74 g/mol.

how many grams of $\ce{ag_{2}cro_{4}}$ form during the reaction?

Explanation:

Step1: Determine mole ratio

From the reaction \(2\text{AgNO}_3(\text{aq}) + \text{K}_2\text{CrO}_4(\text{aq})
ightarrow \text{Ag}_2\text{CrO}_4(\text{s}) + 2\text{KNO}_3(\text{aq})\), the mole ratio of \(\text{AgNO}_3\) to \(\text{Ag}_2\text{CrO}_4\) is \(2:1\). So moles of \(\text{Ag}_2\text{CrO}_4\) = moles of \(\text{AgNO}_3 \times \frac{1}{2}\).
Given moles of \(\text{AgNO}_3 = 0.0750\) mol, moles of \(\text{Ag}_2\text{CrO}_4 = 0.0750 \times \frac{1}{2} = 0.0375\) mol.

Step2: Calculate mass of \(\text{Ag}_2\text{CrO}_4\)

Mass = moles × molar mass. Molar mass of \(\text{Ag}_2\text{CrO}_4\) is \(331.74\) g/mol.
Mass = \(0.0375\) mol × \(331.74\) g/mol = \(12.44\) g (rounded appropriately).

Answer:

\(12.44\) grams (or more precisely, \(0.0375\times331.74 = 12.44025\) which can be rounded to \(12.4\) or \(12.44\) depending on significant figures)