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To determine how many grams of water are theoretically produced in the reaction where we have 2.6 moles of [tex]\( \text{HCl} \)[/tex] and 1.8 moles of [tex]\( \text{Ca(OH)}_2 \)[/tex], we need to follow these steps:
1. Understand the balanced chemical equation:
[tex]\[ \text{2 HCl + }\text{ Ca(OH)}_2 \rightarrow \text{2 H}_2\text{O + CaCl}_2 \][/tex]
This equation tells us that 2 moles of [tex]\( \text{HCl} \)[/tex] react with 1 mole of [tex]\( \text{Ca(OH)}_2 \)[/tex] to produce 2 moles of [tex]\( \text{H}_2\text{O} \)[/tex].
2. Determine the moles of [tex]\( \text{H}_2\text{O} \)[/tex] produced from each reactant:
- From HCl:
According to the balanced equation, 2 moles of [tex]\( \text{HCl} \)[/tex] produce 2 moles of [tex]\( \text{H}_2\text{O} \)[/tex]. Therefore, the number of moles of [tex]\( \text{H}_2\text{O} \)[/tex] produced by [tex]\( \text{HCl} \)[/tex] is:
[tex]\[ 2.6 \text{ moles of HCl} \times \frac{2 \text{ moles of H}_2\text{O}}{2 \text{ moles of HCl}} = 2.6 \text{ moles of } \text{H}_2\text{O} \][/tex]
- From [tex]\( \text{Ca(OH)}_2 \)[/tex]:
According to the balanced equation, 1 mole of [tex]\( \text{Ca(OH)}_2 \)[/tex] produces 2 moles of [tex]\( \text{H}_2\text{O} \)[/tex]. So, the number of moles of [tex]\( \text{H}_2\text{O} \)[/tex] produced by [tex]\( \text{Ca(OH)}_2 \)[/tex] is:
[tex]\[ 1.8 \text{ moles of } \text{Ca(OH)}_2 \times \frac{2 \text{ moles of H}_2\text{O}}{1 \text{ mole of Ca(OH)}_2} = 3.6 \text{ moles of } \text{H}_2\text{O} \][/tex]
3. Determine the limiting reactant:
The limiting reactant is the one that produces the lesser amount of product. Here, [tex]\( \text{HCl} \)[/tex] can produce 2.6 moles of [tex]\( \text{H}_2\text{O} \)[/tex], whereas [tex]\( \text{Ca(OH)}_2 \)[/tex] can produce 3.6 moles of [tex]\( \text{H}_2\text{O} \)[/tex]. Therefore, [tex]\( \text{HCl} \)[/tex] is the limiting reactant because it produces fewer moles of [tex]\( \text{H}_2\text{O} \)[/tex].
4. Calculate the grams of water produced:
The moles of [tex]\( \text{H}_2\text{O} \)[/tex] produced will be limited by the limiting reactant [tex]\( \text{HCl} \)[/tex], which is 2.6 moles of [tex]\( \text{H}_2\text{O} \)[/tex].
The molar mass of [tex]\( \text{H}_2\text{O} \)[/tex] is approximately 18 g/mol (since [tex]\( 2 \times 1 \,\text{g/mol (from hydrogen)} + 1 \times 16 \,\text{g/mol (from oxygen)} = 18 \,\text{g/mol} \)[/tex]).
So, the mass of [tex]\( \text{H}_2\text{O} \)[/tex] produced is:
[tex]\[ 2.6 \text{ moles of } \text{H}_2\text{O} \times 18 \text{ g/mol} = 46.8 \text{ grams of } \text{H}_2\text{O} \][/tex]
Therefore, the amount of water theoretically produced is 46.8 grams. Hence, the correct answer is:
[tex]\[ \boxed{46.8} \][/tex]
1. Understand the balanced chemical equation:
[tex]\[ \text{2 HCl + }\text{ Ca(OH)}_2 \rightarrow \text{2 H}_2\text{O + CaCl}_2 \][/tex]
This equation tells us that 2 moles of [tex]\( \text{HCl} \)[/tex] react with 1 mole of [tex]\( \text{Ca(OH)}_2 \)[/tex] to produce 2 moles of [tex]\( \text{H}_2\text{O} \)[/tex].
2. Determine the moles of [tex]\( \text{H}_2\text{O} \)[/tex] produced from each reactant:
- From HCl:
According to the balanced equation, 2 moles of [tex]\( \text{HCl} \)[/tex] produce 2 moles of [tex]\( \text{H}_2\text{O} \)[/tex]. Therefore, the number of moles of [tex]\( \text{H}_2\text{O} \)[/tex] produced by [tex]\( \text{HCl} \)[/tex] is:
[tex]\[ 2.6 \text{ moles of HCl} \times \frac{2 \text{ moles of H}_2\text{O}}{2 \text{ moles of HCl}} = 2.6 \text{ moles of } \text{H}_2\text{O} \][/tex]
- From [tex]\( \text{Ca(OH)}_2 \)[/tex]:
According to the balanced equation, 1 mole of [tex]\( \text{Ca(OH)}_2 \)[/tex] produces 2 moles of [tex]\( \text{H}_2\text{O} \)[/tex]. So, the number of moles of [tex]\( \text{H}_2\text{O} \)[/tex] produced by [tex]\( \text{Ca(OH)}_2 \)[/tex] is:
[tex]\[ 1.8 \text{ moles of } \text{Ca(OH)}_2 \times \frac{2 \text{ moles of H}_2\text{O}}{1 \text{ mole of Ca(OH)}_2} = 3.6 \text{ moles of } \text{H}_2\text{O} \][/tex]
3. Determine the limiting reactant:
The limiting reactant is the one that produces the lesser amount of product. Here, [tex]\( \text{HCl} \)[/tex] can produce 2.6 moles of [tex]\( \text{H}_2\text{O} \)[/tex], whereas [tex]\( \text{Ca(OH)}_2 \)[/tex] can produce 3.6 moles of [tex]\( \text{H}_2\text{O} \)[/tex]. Therefore, [tex]\( \text{HCl} \)[/tex] is the limiting reactant because it produces fewer moles of [tex]\( \text{H}_2\text{O} \)[/tex].
4. Calculate the grams of water produced:
The moles of [tex]\( \text{H}_2\text{O} \)[/tex] produced will be limited by the limiting reactant [tex]\( \text{HCl} \)[/tex], which is 2.6 moles of [tex]\( \text{H}_2\text{O} \)[/tex].
The molar mass of [tex]\( \text{H}_2\text{O} \)[/tex] is approximately 18 g/mol (since [tex]\( 2 \times 1 \,\text{g/mol (from hydrogen)} + 1 \times 16 \,\text{g/mol (from oxygen)} = 18 \,\text{g/mol} \)[/tex]).
So, the mass of [tex]\( \text{H}_2\text{O} \)[/tex] produced is:
[tex]\[ 2.6 \text{ moles of } \text{H}_2\text{O} \times 18 \text{ g/mol} = 46.8 \text{ grams of } \text{H}_2\text{O} \][/tex]
Therefore, the amount of water theoretically produced is 46.8 grams. Hence, the correct answer is:
[tex]\[ \boxed{46.8} \][/tex]
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