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Sagot :
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]
### 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]
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