Answer: 1. AgNO3
Explanation:
Hi! In order to solve this problem, we must compare the mole values of AgNO3 and CaCl2. Since AgNO3 has a coefficient of 2 and CaCl2 has a coefficient of 1, so AgNO3 has to at least be double the moles of CaCl2 to keep the reaction at balance. If CaCl2 is more than half the moles of AgNO3 it is in excess.
Solving:
[tex]\text{AgNO}_3 (\text{silver nitrate}): & \quad \\ \text{Ag} = 107.87 \ \text{g/mol}\\ \quad \text{N} = 14.01 \ \text{g/mol} \\ \quad 3 \times \text{O} = 3 \times 16.00 \, \text{g/mol} \\& \quad \text{Molar mass of AgNO}_3 = 107.87 + 14.01 + 48.00 = \boxed{169.88} \, \text{g/mol} \\[/tex]
[tex]\text{CaCl}_2 (\text{calcium chloride}): \\ \quad \text{Ca} = 40.08 \ \text{g/mol}\\ \quad 2 \times \text{Cl} = 2 \times 35.45 \, \text{g/mol} \\ \quad \text{Molar mass of CaCl}_2 = 40.08 + 2 \times 35.45 = \boxed{110.98} \, \text{g/mol}[/tex]
Now that we have their molar masses, lets divide the gram amounts given in the question by their molar masses:
[tex]AgNO$_3$:\[\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{80.00 \, \text{g}}{169.88 \, \text{g/mol}} \approx 0.471 \, \text{moles}\][/tex]
[tex]CaCl$_2$:\[\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{30.00 \, \text{g}}{110.98 \, \text{g/mol}} \approx 0.270 \, \text{moles}\][/tex]
Now we can multiply the moles of CaCl2 to see if it is greater than or less than the moles of AgNO3. Less means that CaCl2 is limiting, and more means that AgNO3 is limiting:
[tex]CaCl_2 : 0.270~moles ~\times 2 ~= \boxed{0.540}~ moles[/tex]
Now compare that to AgNO3:
[tex]CaCl_2 : 0.540 ~moles~ \boxed > ~AgNO_3 : 0.471 ~moles[/tex]
Since CaCl2 is greater, AgNO3 is the limiting reactant.
That's it!
