Get expert insights and reliable answers to your questions on IDNLearn.com. Find reliable solutions to your questions quickly and accurately with help from our dedicated community of experts.
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]
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Thank you for choosing IDNLearn.com. We’re committed to providing accurate answers, so visit us again soon.
