1. n₂ + 3h₂ → 2nh₃
how many grams of nh₃ can be produced from the reaction of 28 g of n₂ and 25 g of h₂?
2. how much of the excess reagent in problem 1 is left over?
3. mg + 2hcl → mgcl₂ + h₂
what volume of hydrogen at stp is produced from the reaction of 50.0 g of mg and the equivalent of 75 g of hcl?
4. how much of the excess reagent in problem 3 is left over?
5. 3agno₃ + na₃po₄ → ag₃po₄ + 3nano₃
silver nitrate and sodium phosphate are reacted in equal amounts of 200. g each. how many grams of silver phosphate are produced?
6. how much of the excess reagent in problem 5 is left?

Question

Accepted Answer

### Problem 1
# Explanation:
## Step1: Calculate moles of reactants
Molar mass of $N_2 = 28\space g/mol$, moles of $N_2=\frac{28\space g}{28\space g/mol} = 1\space mol$
Molar mass of $H_2=2\space g/mol$, moles of $H_2=\frac{25\space g}{2\space g/mol}=12.5\space mol$
From the equation $N_2 + 3H_2
ightarrow2NH_3$, the mole ratio of $N_2:H_2 = 1:3$
For $1\space mol$ of $N_2$, $3\space mol$ of $H_2$ is required. Since $H_2$ is in excess ($12.5\space mol>3\space mol$), $N_2$ is the limiting reagent.
## Step2: Calculate moles of $NH_3$
From the mole ratio $N_2:NH_3 = 1:2$, moles of $NH_3 = 2\times1\space mol= 2\space mol$
Molar mass of $NH_3=17\space g/mol$, mass of $NH_3=2\space mol\times17\space g/mol = 34\space g$

# Answer:
$34\space g$

### Problem 2
# Explanation:
## Step1: Calculate moles of $H_2$ reacted
From Problem 1, moles of $N_2 = 1\space mol$. Mole ratio $N_2:H_2 = 1:3$, moles of $H_2$ reacted $=3\times1\space mol = 3\space mol$
## Step2: Calculate mass of $H_2$ left
Initial moles of $H_2 = 12.5\space mol$, moles of $H_2$ left $=12.5 - 3=9.5\space mol$
Mass of $H_2$ left $=9.5\space mol\times2\space g/mol=19\space g$

# Answer:
$19\space g$

### Problem 3
# Explanation:
## Step1: Calculate moles of reactants
Molar mass of $Mg = 24.3\space g/mol$, moles of $Mg=\frac{50.0\space g}{24.3\space g/mol}\approx2.06\space mol$
Molar mass of $HCl = 36.5\space g/mol$, moles of $HCl=\frac{75\space g}{36.5\space g/mol}\approx2.05\space mol$
From the equation $Mg + 2HCl
ightarrow MgCl_2+H_2$, mole ratio $Mg:HCl = 1:2$
For $2.06\space mol$ of $Mg$, $4.12\space mol$ of $HCl$ is required. Since $HCl$ ($2.05\space mol$) is the limiting reagent.
## Step2: Calculate moles of $H_2$
Mole ratio $HCl:H_2=2:1$, moles of $H_2=\frac{2.05\space mol}{2}=1.025\space mol$
At STP ($T = 273\space K$, $P = 1\space atm$), $V=nRT/P$, $R = 0.0821\space L\cdot atm/(mol\cdot K)$
$V=\frac{1.025\space mol\times0.0821\space L\cdot atm/(mol\cdot K)\times273\space K}{1\space atm}\approx22.9\space L$

# Answer:
$22.9\space L$

### Problem 4
# Explanation:
## Step1: Calculate moles of $Mg$ reacted
Mole ratio $HCl:Mg = 2:1$, moles of $Mg$ reacted $=\frac{2.05\space mol}{2}=1.025\space mol$
## Step2: Calculate mass of $Mg$ left
Initial moles of $Mg\approx2.06\space mol$, moles of $Mg$ left $=2.06 - 1.025 = 1.035\space mol$
Mass of $Mg$ left $=1.035\space mol\times24.3\space g/mol\approx25.2\space g$

# Answer:
$25.2\space g$

### Problem 5
# Explanation:
## Step1: Calculate moles of reactants
Molar mass of $AgNO_3=169.9\space g/mol$, moles of $AgNO_3=\frac{200\space g}{169.9\space g/mol}\approx1.18\space mol$
Molar mass of $Na_3PO_4 = 164\space g/mol$, moles of $Na_3PO_4=\frac{200\space g}{164\space g/mol}\approx1.22\space mol$
From the equation $3AgNO_3+Na_3PO_4
ightarrow Ag_3PO_4 + 3NaNO_3$, mole ratio $AgNO_3:Na_3PO_4=3:1$
For $1.18\space mol$ of $AgNO_3$, moles of $Na_3PO_4$ required $=\frac{1.18\space mol}{3}\approx0.39\space mol$. $AgNO_3$ is the limiting reagent.
## Step2: Calculate moles of $Ag_3PO_4$
Mole ratio $AgNO_3:Ag_3PO_4 = 3:1$, moles of $Ag_3PO_4=\frac{1.18\space mol}{3}\approx0.39\space mol$
Molar mass of $Ag_3PO_4 = 418.7\space g/mol$, mass of $Ag_3PO_4=0.39\space mol\times418.7\space g/mol\approx163\space g$

# Answer:
$163\space g$

### Problem 6
# Explanation:
## Step1: Calculate moles of $Na_3PO_4$ reacted
Mole ratio $AgNO_3:Na_3PO_4 = 3:1$, moles of $Na_3PO_4$ reacted $=\frac{1.18\space mol}{3}\approx0.39\space mol$
## Step2: Calculate mass of $Na_3PO_4$ left
Initial moles of $Na_3PO_4\approx1.22\space mol$, moles of $Na_3PO_4$ left $=1.22 - 0.39 = 0.83\space mol$
Mass of $Na_3PO_4$ left $=0.83\space mol\times164\space g/mol\approx136\space g$

# Answer:
$136\space g$

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QUESTION IMAGE

Question

Explanation:

Problem 1

Step1: Calculate moles of reactants

Step2: Calculate moles of \(NH_3\)

Step1: Calculate moles of \(H_2\) reacted

Step2: Calculate mass of \(H_2\) left

Step1: Calculate moles of reactants

Step2: Calculate moles of \(H_2\)

Answer:

Problem 2