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Sagot :
Let's solve the problem step by step.
Step 1: Identify the given information and relevant chemical equation.
- The balanced chemical equation is:
[tex]\( Mg_3N_2(s) + 6 H_2O(l) \rightarrow 3 Mg(OH)_2(s) + 2 NH_3(g) \)[/tex]
- We are given that 150 grams of [tex]\( Mg(OH)_2 \)[/tex] are produced.
Step 2: Determine the molar masses.
- Molar mass of [tex]\( Mg(OH)_2 \)[/tex]:
[tex]\[ \text{Molar mass of } Mg(OH)_2 = Mg + 2 \times (O + H) \][/tex]
[tex]\[ Mg = 24.305 \text{ grams/mole}, O = 15.999 \text{ grams/mole}, H = 1.008 \text{ grams/mole} \][/tex]
[tex]\[ \text{Molar mass of } Mg(OH)_2 = 24.305 + 2 \times (15.999 + 1.008) = 24.305 + 2 \times 17.007 = 24.305 + 34.014 = 58.319 \text{ grams/mole} \][/tex]
- Molar mass of [tex]\( H_2O \)[/tex]:
[tex]\[ \text{Molar mass of } H_2O = 2 \times H + O = 2 \times 1.008 + 15.999 = 2.016 + 15.999 = 18.015 \text{ grams/mole} \][/tex]
Step 3: Calculate the number of moles of [tex]\( Mg(OH)_2 \)[/tex] produced.
- Using the molar mass of [tex]\( Mg(OH)_2 \)[/tex]:
[tex]\[ \text{Moles of } Mg(OH)_2 = \frac{150 \text{ grams}}{58.319 \text{ grams/mole}} \approx 2.572 \text{ moles} \][/tex]
Step 4: Determine the stoichiometric relationship between [tex]\( Mg(OH)_2 \)[/tex] and [tex]\( H_2O \)[/tex].
- From the balanced equation: [tex]\( 3 \text{ moles of } Mg(OH)_2 \)[/tex] are produced from [tex]\( 6 \text{ moles of } H_2O \)[/tex].
- Therefore, for 1 mole of [tex]\( Mg(OH)_2 \)[/tex], there is a requirement of [tex]\( 2 \text{ moles of } H_2O \)[/tex].
Step 5: Calculate the number of moles of [tex]\( H_2O \)[/tex] needed.
- Using the stoichiometric ratio from the balanced equation:
[tex]\[ \text{Moles of } H_2O = \left( \frac{6}{3} \right) \times \text{moles of } Mg(OH)_2 = 2 \times 2.572 \approx 5.144 \text{ moles} \][/tex]
Step 6: Calculate the mass of [tex]\( H_2O \)[/tex] needed.
- Using the molar mass of [tex]\( H_2O \)[/tex]:
[tex]\[ \text{Grams of } H_2O = \text{moles of } H_2O \times \text{molar mass of } H_2O = 5.144 \times 18.015 \approx 92.671 \text{ grams} \][/tex]
Therefore, the number of grams of [tex]\( H_2O \)[/tex] needed to produce 150 grams of [tex]\( Mg(OH)_2 \)[/tex] is approximately 93 grams.
Step 1: Identify the given information and relevant chemical equation.
- The balanced chemical equation is:
[tex]\( Mg_3N_2(s) + 6 H_2O(l) \rightarrow 3 Mg(OH)_2(s) + 2 NH_3(g) \)[/tex]
- We are given that 150 grams of [tex]\( Mg(OH)_2 \)[/tex] are produced.
Step 2: Determine the molar masses.
- Molar mass of [tex]\( Mg(OH)_2 \)[/tex]:
[tex]\[ \text{Molar mass of } Mg(OH)_2 = Mg + 2 \times (O + H) \][/tex]
[tex]\[ Mg = 24.305 \text{ grams/mole}, O = 15.999 \text{ grams/mole}, H = 1.008 \text{ grams/mole} \][/tex]
[tex]\[ \text{Molar mass of } Mg(OH)_2 = 24.305 + 2 \times (15.999 + 1.008) = 24.305 + 2 \times 17.007 = 24.305 + 34.014 = 58.319 \text{ grams/mole} \][/tex]
- Molar mass of [tex]\( H_2O \)[/tex]:
[tex]\[ \text{Molar mass of } H_2O = 2 \times H + O = 2 \times 1.008 + 15.999 = 2.016 + 15.999 = 18.015 \text{ grams/mole} \][/tex]
Step 3: Calculate the number of moles of [tex]\( Mg(OH)_2 \)[/tex] produced.
- Using the molar mass of [tex]\( Mg(OH)_2 \)[/tex]:
[tex]\[ \text{Moles of } Mg(OH)_2 = \frac{150 \text{ grams}}{58.319 \text{ grams/mole}} \approx 2.572 \text{ moles} \][/tex]
Step 4: Determine the stoichiometric relationship between [tex]\( Mg(OH)_2 \)[/tex] and [tex]\( H_2O \)[/tex].
- From the balanced equation: [tex]\( 3 \text{ moles of } Mg(OH)_2 \)[/tex] are produced from [tex]\( 6 \text{ moles of } H_2O \)[/tex].
- Therefore, for 1 mole of [tex]\( Mg(OH)_2 \)[/tex], there is a requirement of [tex]\( 2 \text{ moles of } H_2O \)[/tex].
Step 5: Calculate the number of moles of [tex]\( H_2O \)[/tex] needed.
- Using the stoichiometric ratio from the balanced equation:
[tex]\[ \text{Moles of } H_2O = \left( \frac{6}{3} \right) \times \text{moles of } Mg(OH)_2 = 2 \times 2.572 \approx 5.144 \text{ moles} \][/tex]
Step 6: Calculate the mass of [tex]\( H_2O \)[/tex] needed.
- Using the molar mass of [tex]\( H_2O \)[/tex]:
[tex]\[ \text{Grams of } H_2O = \text{moles of } H_2O \times \text{molar mass of } H_2O = 5.144 \times 18.015 \approx 92.671 \text{ grams} \][/tex]
Therefore, the number of grams of [tex]\( H_2O \)[/tex] needed to produce 150 grams of [tex]\( Mg(OH)_2 \)[/tex] is approximately 93 grams.
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