Question

cu(s) + 4 hno₃(aq) → cu(no₃)₂(aq) + 2 no₂(g) + 2 h₂o(l)
a student analyzes a 2.00 g sample of a mixture of copper, cu, and aluminum, al, by reacting the copper with nitric acid, hno₃ as represented in the equation above. the student determines that the reaction produces 0.010 mol cu(no₃)₂. assuming that all of the copper in the mixture reacted completely, what was the percent of cu by mass in the 2.00g sample of the mixture?
gravimetric factor of cu(no₃)₂ atomic mass of cu
? g cu = 0.010 mol cu(no₃)₂ x \frac{1 mol cu}{1 mol cu(no₃)₂} x \frac{64 g cu}{1 mol cu}=0.64 g cu
% cu=\frac{mass cu}{mass sample}=\frac{0.64 g cu}{2.00 g mixture}= 32% cu

1. a 3.0 g sample of a mixture of calcium chloride, cacl₂, and sodium chloride, kcl, is found to contain 0.18 g cl. what percent of this sample is cacl₂?

atomic mass of cl gravimetric factor of cacl₂ molar mass of cacl₂
? g cacl₂ = 0.18 g cl x 35.45g/mol x \frac{1mol cl}{1mol cl (in cacl₂)}x \frac{70.9 g cacl₂}{1mol cacl₂}=2.26 or 0.28g
% of compound=\frac{mass of the compound}{mass of the sample mixture}x 100%=\frac{2.26g cacl₂}{3.0g}x 100% = 9.3%

1. calculate the molarity of the solution made by dissolving 25.5 g bacl₂ in enough water to form 355 ml of the solution.

molarity (m)=\frac{moles of solute}{l of solution}

1. how many moles of naoh are there in 25 ml of 0.75 m naoh?

? mol of naoh = 0.75 mol naoh/l x 25 ml/l = 0.03 mol

Explanation:

Step1: Calculate moles of $CaCl_2$ from moles of $Cl$

The atomic mass of $Cl$ is approximately $35.45\ g/mol$. First, find moles of $Cl$ in the sample: $n_{Cl}=\frac{0.18\ g}{35.45\ g/mol}\approx0.00508\ mol$. In $CaCl_2$, there are 2 moles of $Cl$ for every 1 mole of $CaCl_2$. So moles of $CaCl_2$, $n_{CaCl_2}=\frac{0.00508\ mol}{2}= 0.00254\ mol$. The molar - mass of $CaCl_2$ is $M_{CaCl_2}=40.08 + 2\times35.45=110.98\ g/mol$. Then the mass of $CaCl_2$, $m_{CaCl_2}=n_{CaCl_2}\times M_{CaCl_2}=0.00254\ mol\times110.98\ g/mol\approx0.282\ g$.

Step2: Calculate percentage of $CaCl_2$ in the sample

The percentage of $CaCl_2$ in the sample is $\%CaCl_2=\frac{m_{CaCl_2}}{m_{sample}}\times100\%=\frac{0.282\ g}{3.0\ g}\times100\%\approx9.4\%$.

Step3: Calculate molarity of $BaCl_2$ solution

The molar - mass of $BaCl_2$ is $M_{BaCl_2}=137.33+2\times35.45 = 208.23\ g/mol$. Moles of $BaCl_2$, $n_{BaCl_2}=\frac{25.5\ g}{208.23\ g/mol}\approx0.122\ mol$. The volume of the solution $V = 355\ mL=0.355\ L$. Molarity $M=\frac{n_{BaCl_2}}{V}=\frac{0.122\ mol}{0.355\ L}\approx0.344\ M$.

Step4: Calculate moles of $NaOH$

The molarity formula is $M=\frac{n}{V}$, where $M = 0.75\ M$ and $V=25\ mL = 0.025\ L$. Rearranging for $n$, we get $n = M\times V$. So $n_{NaOH}=0.75\ mol/L\times0.025\ L = 0.01875\ mol$.

Answer:

1. Approximately $9.4\%$
2. Approximately $0.344\ M$
3. $0.01875\ mol$