Question

1. molarity, abbreviated m, is defined as the concentration of a solution expressed as moles of solute per liter of solution. how many milliliters of 4.00 m h₂so₄ are required to react with 1.60 g of cuo according to equation 4? 6. if 2.00 g of zn is allowed to react with 2.00 g of cuso₄, according to equation 5, how many grams of zn will remain after the reaction is complete?

Explanation:

Step1: Write the chemical reaction equations

For the reaction of $H_2SO_4$ and $CuO$: $H_2SO_4+CuO = CuSO_4 + H_2O$. For the reaction of $Zn$ and $CuSO_4$: $Zn+CuSO_4=ZnSO_4 + Cu$.

Step2: Calculate moles of $CuO$ in question 5

The molar - mass of $CuO$ is $M_{CuO}=63.5 + 16=79.5\ g/mol$. The number of moles of $CuO$, $n_{CuO}=\frac{m_{CuO}}{M_{CuO}}=\frac{1.60\ g}{79.5\ g/mol}\approx0.0201\ mol$.
From the reaction equation $H_2SO_4+CuO = CuSO_4 + H_2O$, the mole - ratio of $H_2SO_4$ to $CuO$ is $1:1$. So the moles of $H_2SO_4$ required, $n_{H_2SO_4}=n_{CuO}=0.0201\ mol$.
We know that molarity $M = \frac{n}{V}$ (where $M$ is molarity, $n$ is moles of solute, and $V$ is volume of solution in liters). Given $M = 4.00\ M$, then $V=\frac{n}{M}=\frac{0.0201\ mol}{4.00\ mol/L}=0.005025\ L$. Converting to milliliters, $V = 5.025\ mL$.

Step3: Calculate moles of $Zn$ and $CuSO_4$ in question 6

The molar - mass of $Zn$ is $M_{Zn}=65.4\ g/mol$, so the number of moles of $Zn$, $n_{Zn}=\frac{m_{Zn}}{M_{Zn}}=\frac{2.00\ g}{65.4\ g/mol}\approx0.0306\ mol$.
The molar - mass of $CuSO_4$ is $M_{CuSO_4}=63.5 + 32+4\times16 = 159.5\ g/mol$. The number of moles of $CuSO_4$, $n_{CuSO_4}=\frac{m_{CuSO_4}}{M_{CuSO_4}}=\frac{2.00\ g}{159.5\ g/mol}\approx0.0125\ mol$.
From the reaction equation $Zn + CuSO_4=ZnSO_4+Cu$, the mole - ratio of $Zn$ to $CuSO_4$ is $1:1$. $CuSO_4$ is the limiting reactant.
The moles of $Zn$ that react with $CuSO_4$ is $n_{Zn - reacted}=n_{CuSO_4}=0.0125\ mol$.
The moles of $Zn$ remaining, $n_{Zn - remaining}=n_{Zn}-n_{Zn - reacted}=0.0306\ mol - 0.0125\ mol = 0.0181\ mol$.
The mass of $Zn$ remaining, $m_{Zn - remaining}=n_{Zn - remaining}\times M_{Zn}=0.0181\ mol\times65.4\ g/mol\approx1.18\ g$.

Answer:

1. $5.025\ mL$
2. $1.18\ g$