Question

we wish to determine how many grams of solid ag₂cro₄ will precipitate when 150. ml of 0.500 m agno₃ solution is added to excess k₂cro₄. 2agno₃(aq) + k₂cro₄(aq) → ag₂cro₄(s) + 2kno₃(aq) how many moles of agno₃ are present in 150. ml of 0.500 m agno₃?

Explanation:

Step1: Recall the formula for moles from molarity and volume

The formula relating moles ($n$), molarity ($M$), and volume ($V$ in liters) is $n = M\times V$. First, we need to convert the volume from milliliters to liters.
Given $V = 150. \, \text{mL}$, we convert it to liters by dividing by 1000: $V = \frac{150.}{1000} \, \text{L} = 0.150 \, \text{L}$. The molarity $M = 0.500 \, \text{M}$.

Step2: Calculate the moles of $\ce{AgNO3}$

Using the formula $n = M\times V$, substitute the values of $M$ and $V$:
$n = 0.500 \, \text{mol/L} \times 0.150 \, \text{L}$

Answer:

$0.0750$ moles (The calculation is $0.500\times0.150 = 0.0750$)