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Sure! Let's work through this example step by step.
First, we need to understand the given details and the chemical reaction involved:
The chemical reaction:
NH[tex]\(_2\)[/tex]Cl + Ca(OH)[tex]\(_2\)[/tex] → CaCl[tex]\(_2\)[/tex] + H[tex]\(_2\)[/tex]O + NH[tex]\(_3\)[/tex]
This equation shows that 1 mole of NH[tex]\(_2\)[/tex]Cl reacts with Ca(OH)[tex]\(_2\)[/tex] to produce 1 mole of CaCl[tex]\(_2\)[/tex].
Step 1: Calculate moles of NH[tex]\(_2\)[/tex]Cl
We are given the mass of NH[tex]\(_2\)[/tex]Cl as 80.25 grams, and we need to use its molar mass to find the moles. The molar mass of NH[tex]\(_2\)[/tex]Cl (calculated by adding up the atomic masses of nitrogen (N), hydrogen (H), and chlorine (Cl)) is 51.5 g/mol.
To find the moles of NH[tex]\(_2\)[/tex]Cl, we use the formula:
[tex]\[ \text{moles of NH\(_2\)Cl} = \frac{\text{mass of NH\(_2\)Cl}}{\text{molar mass of NH\(_2\)Cl}} \][/tex]
Plugging in the values, we get:
[tex]\[ \text{moles of NH\(_2\)Cl} = \frac{80.25 \, \text{grams}}{51.5 \, \text{g/mol}} = 1.558252427184466 \, \text{moles} \][/tex]
Step 2: Calculate moles of CaCl[tex]\(_2\)[/tex] produced
From the stoichiometry of the balanced equation, we see that 1 mole of NH[tex]\(_2\)[/tex]Cl produces 1 mole of CaCl[tex]\(_2\)[/tex]. Therefore, the moles of CaCl[tex]\(_2\)[/tex] produced will be the same as the moles of NH[tex]\(_2\)[/tex]Cl reacted.
So, the moles of CaCl[tex]\(_2\)[/tex] produced is also 1.558252427184466 moles.
Step 3: Calculate the mass of CaCl[tex]\(_2\)[/tex] produced
To find the mass of CaCl[tex]\(_2\)[/tex], we use the molar mass of CaCl[tex]\(_2\)[/tex]. The molar mass of CaCl[tex]\(_2\)[/tex] (calculated by adding the atomic masses of calcium (Ca) and chlorine (Cl)) is 110.98 g/mol.
The formula to find the mass is:
[tex]\[ \text{mass of CaCl\(_2\)} = \text{moles of CaCl\(_2\)} \times \text{molar mass of CaCl\(_2\)} \][/tex]
Plugging in the values, we get:
[tex]\[ \text{mass of CaCl\(_2\)} = 1.558252427184466 \, \text{moles} \times 110.98 \, \text{g/mol} = 172.93485436893204 \, \text{grams} \][/tex]
Therefore, if 80.25 grams of NH[tex]\(_2\)[/tex]Cl reacts with an excess of Ca(OH)[tex]\(_2\)[/tex], the mass of CaCl[tex]\(_2\)[/tex] that will form is 172.93 grams (rounded to two decimal places).
First, we need to understand the given details and the chemical reaction involved:
The chemical reaction:
NH[tex]\(_2\)[/tex]Cl + Ca(OH)[tex]\(_2\)[/tex] → CaCl[tex]\(_2\)[/tex] + H[tex]\(_2\)[/tex]O + NH[tex]\(_3\)[/tex]
This equation shows that 1 mole of NH[tex]\(_2\)[/tex]Cl reacts with Ca(OH)[tex]\(_2\)[/tex] to produce 1 mole of CaCl[tex]\(_2\)[/tex].
Step 1: Calculate moles of NH[tex]\(_2\)[/tex]Cl
We are given the mass of NH[tex]\(_2\)[/tex]Cl as 80.25 grams, and we need to use its molar mass to find the moles. The molar mass of NH[tex]\(_2\)[/tex]Cl (calculated by adding up the atomic masses of nitrogen (N), hydrogen (H), and chlorine (Cl)) is 51.5 g/mol.
To find the moles of NH[tex]\(_2\)[/tex]Cl, we use the formula:
[tex]\[ \text{moles of NH\(_2\)Cl} = \frac{\text{mass of NH\(_2\)Cl}}{\text{molar mass of NH\(_2\)Cl}} \][/tex]
Plugging in the values, we get:
[tex]\[ \text{moles of NH\(_2\)Cl} = \frac{80.25 \, \text{grams}}{51.5 \, \text{g/mol}} = 1.558252427184466 \, \text{moles} \][/tex]
Step 2: Calculate moles of CaCl[tex]\(_2\)[/tex] produced
From the stoichiometry of the balanced equation, we see that 1 mole of NH[tex]\(_2\)[/tex]Cl produces 1 mole of CaCl[tex]\(_2\)[/tex]. Therefore, the moles of CaCl[tex]\(_2\)[/tex] produced will be the same as the moles of NH[tex]\(_2\)[/tex]Cl reacted.
So, the moles of CaCl[tex]\(_2\)[/tex] produced is also 1.558252427184466 moles.
Step 3: Calculate the mass of CaCl[tex]\(_2\)[/tex] produced
To find the mass of CaCl[tex]\(_2\)[/tex], we use the molar mass of CaCl[tex]\(_2\)[/tex]. The molar mass of CaCl[tex]\(_2\)[/tex] (calculated by adding the atomic masses of calcium (Ca) and chlorine (Cl)) is 110.98 g/mol.
The formula to find the mass is:
[tex]\[ \text{mass of CaCl\(_2\)} = \text{moles of CaCl\(_2\)} \times \text{molar mass of CaCl\(_2\)} \][/tex]
Plugging in the values, we get:
[tex]\[ \text{mass of CaCl\(_2\)} = 1.558252427184466 \, \text{moles} \times 110.98 \, \text{g/mol} = 172.93485436893204 \, \text{grams} \][/tex]
Therefore, if 80.25 grams of NH[tex]\(_2\)[/tex]Cl reacts with an excess of Ca(OH)[tex]\(_2\)[/tex], the mass of CaCl[tex]\(_2\)[/tex] that will form is 172.93 grams (rounded to two decimal places).
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