Answered

When magnesium carbonate is added to nitric acid, magnesium nitrate, carbon dioxide, and water are produced.

[tex]\[ MgCO_3(s) + 2 HNO_3(aq) \longrightarrow Mg(NO_3)_2(aq) + H_2O(l) + CO_2(g) \][/tex]

1. How many grams of magnesium nitrate will be produced in the reaction when 31.0 g of magnesium carbonate is combined with 12.0 g of nitric acid?
- Mass of [tex]\( Mg(NO_3)_2 \)[/tex]: [tex]\(\square\)[/tex] g

2. How many grams of magnesium carbonate remain after the reaction is complete?
- Mass of [tex]\( MgCO_3 \)[/tex]: [tex]\(\square\)[/tex] g

3. How many grams of nitric acid remain after the reaction is complete?
- Mass of [tex]\( HNO_3 \)[/tex]: [tex]\(\square\)[/tex] g



Answer :

GinnyAnswer
Certainly! Let's solve this problem step-by-step:

### Given Data:
- Mass of magnesium carbonate ([tex]\(MgCO_3\)[/tex]) = 31.0 g
- Mass of nitric acid ([tex]\(HNO_3\)[/tex]) = 12.0 g

### Step 1: Calculate the number of moles of each reactant

#### Magnesium Carbonate ([tex]\(MgCO_3\)[/tex]):
- Molecular weight of [tex]\(MgCO_3\)[/tex]: [tex]\(24.305\)[/tex] (Mg) + [tex]\(12.01\)[/tex] (C) + [tex]\(3 \times 16.00\)[/tex] (O) = [tex]\(84.315 \, \text{g/mol}\)[/tex]

[tex]\[ \text{Moles} \, \text{of} \, MgCO_3 = \frac{31.0 \, \text{g}}{84.315 \, \text{g/mol}} \approx 0.3675 \, \text{moles} \][/tex]

#### Nitric Acid ([tex]\(HNO_3\)[/tex]):
- Molecular weight of [tex]\(HNO_3\)[/tex]: [tex]\(1.01\)[/tex] (H) + [tex]\(14.01\)[/tex] (N) + [tex]\(3 \times 16.00\)[/tex] (O) = [tex]\(63.01 \, \text{g/mol}\)[/tex]

[tex]\[ \text{Moles} \, \text{of} \, HNO_3 = \frac{12.0 \, \text{g}}{63.01 \, \text{g/mol}} \approx 0.1905 \, \text{moles} \][/tex]

### Step 2: Determine the limiting reagent
Based on the stoichiometry: [tex]\(1 \, \text{mole of } \, MgCO_3\)[/tex] reacts with [tex]\(2 \, \text{moles of } \, HNO_3\)[/tex].

[tex]\[ 0.3675 \, \text{moles of } \, MgCO_3 \, \text{requires } \, 0.3675 \times 2 = 0.735 \, \text{moles of } \, HNO_3 \][/tex]

We only have [tex]\(0.1905 \, \text{moles of } \, HNO_3\)[/tex], which means [tex]\(HNO_3\)[/tex] is the limiting reagent.

### Step 3: Calculate the amount of magnesium carbonate used and the products formed

[tex]\[ \text{Moles} \, \text{of } \, HNO_3 \, \text{used} = 0.1905 \, \text{moles} \, \text{(all of it)} \][/tex]

According to the reaction stoichiometry:
[tex]\[ 1 \, \text{mole of } \, MgCO_3 \, \text{reacts with } \, 2 \, \text{moles of } \, HNO_3 \][/tex]

[tex]\[ \text{Moles of } \, MgCO_3 \, \text{reacted} = \frac{0.1905 \, \text{moles of } \, HNO_3}{2} = 0.09525 \, \text{moles} \][/tex]

### Step 4: Calculate the mass of magnesium nitrate produced

- Molecular weight of [tex]\(Mg(NO_3)_2\)[/tex]: [tex]\(24.305\)[/tex] (Mg) + [tex]\(2 \times (14.01 + 3 \times 16.00)\)[/tex] (NO3 groups) = [tex]\(148.31 \, \text{g/mol}\)[/tex]

[tex]\[ \text{Moles of magnesium nitrate } = 0.09525 \, \text{moles} \, \text{(since it's 1:1 with } \, MgCO_3) \][/tex]

[tex]\[ \text{Mass of magnesium nitrate} = 0.09525 \, \text{moles} \times 148.31 \, \text{g/mol} = 14.1217 \, \text{g} \][/tex]

### Step 5: Calculate the remaining mass of magnesium carbonate and nitric acid

#### Remaining magnesium carbonate:
[tex]\[ \text{Moles of } \, MgCO_3 \, \text{remaining} = 0.3675 \, \text{moles} - 0.09525 \, \text{moles} = 0.27225 \, \text{moles} \][/tex]

[tex]\[ \text{Mass of } \, MgCO_3 \, \text{remaining} = 0.27225 \, \text{moles} \times 84.315 \, \text{g/mol} = 22.9725 \, \text{g} \][/tex]

#### Remaining nitric acid:
Since [tex]\(HNO_3\)[/tex] is the limiting reagent, there is no [tex]\(HNO_3\)[/tex] left.

[tex]\[ \text{Mass of } \, HNO_3 \, \text{remaining} = 0 \, \text{g} \][/tex]

### Final Answers:

1. Mass of magnesium nitrate produced: [tex]\(\boxed{14.1217 \, \text{g}}\)[/tex]
2. Mass of magnesium carbonate remaining: [tex]\(\boxed{22.9725 \, \text{g}}\)[/tex]
3. Mass of nitric acid remaining: [tex]\(\boxed{0.0 \, \text{g}}\)[/tex]