stoichiometry test\nmake sure to show all wor...
stoichiometry test\nmake sure to show all work including conversion factors and give correct units with your answers!\n1. 2 fecl₃ + 3 mgo → fe₂o₃ + 3 mgcl₂\nhow many grams of mgcl₂ are produced from the reaction of 9.5 moles fecl₃?\n2. 2 zns + 3o₂ → 2 zno + 2 so₂\nhow many grams of so₂ are produced from 136.7 grams of oxygen gas?\n3. ____ nh₃ → ____n₂ + ____ h₂\nhow many liters of n₂ gas are produced from 4.8 x 10²³ molecules of nh₃?\n4. ____p₄ + ____o₂ → ____ p₂o₃\nhow many molecules of oxygen are needed to make 11.0 grams of p₂o₃?\n5. 2 co + o₂ → 2co₂\nhow many grams of carbon dioxide will be made from 3.2 x 10²² molecules of o₂?
Answer
### 1.
# Explanation:
## Step1: Determine mole - ratio
From the balanced equation $2FeCl_3 + 3MgO\rightarrow Fe_2O_3+3MgCl_2$, the mole - ratio of $FeCl_3$ to $MgCl_2$ is $2:3$.
## Step2: Calculate moles of $MgCl_2$
If we have $n(FeCl_3)=9.5$ moles, then $n(MgCl_2)=\frac{3}{2}\times n(FeCl_3)=\frac{3}{2}\times9.5 = 14.25$ moles.
## Step3: Calculate mass of $MgCl_2$
The molar mass of $MgCl_2$ is $M(MgCl_2)=24.31+(2\times35.45)=95.21$ g/mol. So, $m(MgCl_2)=n(MgCl_2)\times M(MgCl_2)=14.25\times95.21 = 1356.74$ g.
# Answer:
$1356.74$ g
### 2.
# Explanation:
## Step1: Calculate moles of $O_2$
The molar mass of $O_2$ is $M(O_2) = 32$ g/mol. Given $m(O_2)=136.7$ g, then $n(O_2)=\frac{m(O_2)}{M(O_2)}=\frac{136.7}{32}=4.272$ moles.
## Step2: Determine mole - ratio
From the balanced equation $2ZnS + 3O_2\rightarrow2ZnO + 2SO_2$, the mole - ratio of $O_2$ to $SO_2$ is $3:2$.
## Step3: Calculate moles of $SO_2$
$n(SO_2)=\frac{2}{3}\times n(O_2)=\frac{2}{3}\times4.272 = 2.848$ moles.
## Step4: Calculate mass of $SO_2$
The molar mass of $SO_2$ is $M(SO_2)=32+(2\times16)=64$ g/mol. So, $m(SO_2)=n(SO_2)\times M(SO_2)=2.848\times64 = 182.27$ g.
# Answer:
$182.27$ g
### 3.
First, balance the equation: $2NH_3\rightarrow N_2 + 3H_2$
# Explanation:
## Step1: Calculate moles of $NH_3$
Using Avogadro's number $N_A = 6.022\times10^{23}$ molecules/mol. Given $N(NH_3)=4.8\times 10^{23}$ molecules, then $n(NH_3)=\frac{N(NH_3)}{N_A}=\frac{4.8\times 10^{23}}{6.022\times10^{23}} = 0.797$ moles.
## Step2: Determine mole - ratio
The mole - ratio of $NH_3$ to $N_2$ is $2:1$. So, $n(N_2)=\frac{1}{2}\times n(NH_3)=\frac{1}{2}\times0.797 = 0.3985$ moles.
## Step3: Calculate volume of $N_2$
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 L. So, $V(N_2)=n(N_2)\times22.4=0.3985\times22.4 = 8.93$ L.
# Answer:
$8.93$ L
### 4.
First, balance the equation: $2P_4+6O_2\rightarrow4P_2O_3$
# Explanation:
## Step1: Calculate moles of $P_2O_3$
The molar mass of $P_2O_3$ is $M(P_2O_3)=(2\times30.97)+(3\times16)=109.94$ g/mol. Given $m(P_2O_3)=11.0$ g, then $n(P_2O_3)=\frac{m(P_2O_3)}{M(P_2O_3)}=\frac{11.0}{109.94}=0.1$ moles.
## Step2: Determine mole - ratio
The mole - ratio of $O_2$ to $P_2O_3$ is $6:4 = 3:2$. So, $n(O_2)=\frac{3}{2}\times n(P_2O_3)=\frac{3}{2}\times0.1 = 0.15$ moles.
## Step3: Calculate number of molecules of $O_2$
Using Avogadro's number, $N(O_2)=n(O_2)\times N_A=0.15\times6.022\times10^{23}=9.033\times 10^{22}$ molecules.
# Answer:
$9.033\times 10^{22}$ molecules
### 5.
# Explanation:
## Step1: Calculate moles of $O_2$
Using Avogadro's number $N_A = 6.022\times10^{23}$ molecules/mol. Given $N(O_2)=3.2\times 10^{22}$ molecules, then $n(O_2)=\frac{N(O_2)}{N_A}=\frac{3.2\times 10^{22}}{6.022\times10^{23}}=0.0531$ moles.
## Step2: Determine mole - ratio
From the balanced equation $2CO + O_2\rightarrow2CO_2$, the mole - ratio of $O_2$ to $CO_2$ is $1:2$. So, $n(CO_2)=2\times n(O_2)=2\times0.0531 = 0.1062$ moles.
## Step3: Calculate mass of $CO_2$
The molar mass of $CO_2$ is $M(CO_2)=12+(2\times16)=44$ g/mol. So, $m(CO_2)=n(CO_2)\times M(CO_2)=0.1062\times44 = 4.67$ g.
# Answer:
$4.67$ g