Question

1. how many moles of nickel(iv) sulfate: ni(so₄)₂ is 120.5 g of nickel(iv) sulfate?
2. how many dicarbon hexabromide: c₂br₆ molecules are in 3.5 kg of dicarbon hexabromide? (3 pts.)
3. manganese(vii) oxide (mn₂o₇) is made by reacting 4 moles of manganese (mn) with 7 moles of oxygen (o₂) to make 2 moles of manganese(vii) oxide: 4mn + 7o₂ -------> 2mn₂o₇

given 125 grams of manganese, how many grams of oxygen will be needed for this reaction? how many grams of manganese(vii) oxide can be produced? (4 pts.)

Explanation:

14.

Step1: Calculate molar mass of $Ni(SO_4)_2$

The molar mass of $Ni$ is approximately $58.69\ g/mol$, $S$ is approximately $32.07\ g/mol$ and $O$ is approximately $16.00\ g/mol$.
$M(Ni(SO_4)_2)=58.69 + 2\times(32.07+4\times16.00)=58.69+2\times(32.07 + 64.00)=58.69+2\times96.07=58.69 + 192.14=250.83\ g/mol$

Step2: Calculate moles of $Ni(SO_4)_2$

Use the formula $n=\frac{m}{M}$, where $m = 120.5\ g$ and $M = 250.83\ g/mol$.
$n=\frac{120.5}{250.83}\approx0.48\ mol$

Step1: Calculate molar mass of $C_2Br_6$

The molar mass of $C$ is approximately $12.01\ g/mol$ and $Br$ is approximately $79.90\ g/mol$.
$M(C_2Br_6)=2\times12.01+6\times79.90=24.02 + 479.4=503.42\ g/mol$

Step2: Calculate moles of $C_2Br_6$

First, convert the mass to grams: $m = 3.5\ kg=3500\ g$. Then use $n=\frac{m}{M}$.
$n=\frac{3500}{503.42}\approx6.95\ mol$

Step3: Calculate number of molecules

Use Avogadro's number $N_A = 6.022\times10^{23}\ molecules/mol$.
$N=n\times N_A=6.95\times6.022\times10^{23}\approx4.19\times 10^{24}\ molecules$

Step1: Calculate moles of $Mn$

The molar mass of $Mn$ is approximately $54.94\ g/mol$. Given $m = 125\ g$, use $n=\frac{m}{M}$.
$n(Mn)=\frac{125}{54.94}\approx2.28\ mol$

Step2: Calculate moles of $O_2$ needed

From the balanced - equation $4Mn + 7O_2
ightarrow2Mn_2O_7$, the mole ratio of $Mn$ to $O_2$ is $\frac{n(O_2)}{n(Mn)}=\frac{7}{4}$.
$n(O_2)=\frac{7}{4}\times n(Mn)=\frac{7}{4}\times2.28 = 3.99\ mol$
The molar mass of $O_2$ is $M(O_2)=2\times16.00 = 32.00\ g/mol$.
$m(O_2)=n(O_2)\times M(O_2)=3.99\times32.00 = 127.68\ g$

Step3: Calculate moles of $Mn_2O_7$ produced

The mole ratio of $Mn$ to $Mn_2O_7$ is $\frac{n(Mn_2O_7)}{n(Mn)}=\frac{2}{4}=\frac{1}{2}$.
$n(Mn_2O_7)=\frac{1}{2}\times n(Mn)=\frac{1}{2}\times2.28 = 1.14\ mol$
The molar mass of $Mn_2O_7$ is $M(Mn_2O_7)=2\times54.94+7\times16.00=109.88 + 112.00=221.88\ g/mol$.
$m(Mn_2O_7)=n(Mn_2O_7)\times M(Mn_2O_7)=1.14\times221.88\approx252.95\ g$

Answer:

$0.48\ mol$

15.