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3 years ago

In the following chemical reaction, which compound is a precipitate? NaCl(aq) + AgNO3(aq)NaNO3(aq) + AgCI (s)

Chemistry

2 answers:

disa [49]3 years ago

8 0

Answer: D: AgCl is the correct answer

hjlf3 years ago

4 0

The answer is: [D]: " AgCl " .
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Explanation:
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Given the chemical reaction:
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NaCl (aq) + AgNO₃ (aq) <span>→ </span> NaNO₃ (aq) + AgCl (s) ;
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The "reactants" , which are on the "left-hand side" of the chemical equation, are:

"NaCl" (in aqueous solution); and "AgNO₃" (in aqueous solution).

The "products", which are on the "right-hand side" of the chemical equation, are:

"NaNO₃" (in aqueous solution); and: "AgCl" (as a "solid").

The "AgCl" (as a "product"—which is a "solid"—would be the "precipitate" ;
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→ which is: Answer choice: [D]: " AgCl " .
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For each of the following unbalanced chemical equations suppose that exactly 50.0 g of each reactant is taken. Determine which r

Helen [10]

Answer:

1) Br2 is the limiting reactant.

Mass NaBr produced = 64.4 grams

2) CuSO4 is the limiting reactant

Mass Cu = 19.89 grams

Mass ZnSO4 = 50.54 grams

3) NH4Cl is the limiting reactant

Mass NaCl = 54.6 grams

Mass NH3 =15.9 grams

Mass H2O =16.8 grams

4) Fe2O3 is the limiting reactant

Mass Fe = 35.0 grams

Mass CO2 = 41.3 grams

Explanation:

1) Na+br2 ------------->Nabr

Step 1: Data given

Mass Na = 50.0 grams

Mass Br2 = 50.0 grams

Molar mass Na = 22.99 g/mol

Molar mass Br2 = 159.81 g/mol

Step 2: The balanced equation

2Na + Br2 → 2NaBr

Step 3: Calculate moles

Moles = mass / molar mass

Moles Na = 50.0 grams / 22.99 g/mol = 2.17 moles

Moles Br2 = 50.0 grams / 159.81 g/mol = 0.313 moles

Step 4: Calculate limiting reactant

Br2 is the limiting reactant. It will completely be consumed (0.313 moles).

Na is in excess. There will react 2*0.313 = 0.626 moles

There will remain 2.17 - 0.626 = 1.544 moles

Step 5: Calculate moles NaBr

For 1 mol Br2 we'll have 2 moles NaBr

For 0.313 moles we'll have 0.626 moles NaBr

Step 6: Calculate mass NaBr

Mass NaBr = 0.626 moles * 102.89 g/mol

Mass NaBr = 64.4 grams

2) Zn+cuso4 -------------->Znso4+Cu

Step 1: Data given

Mass Zn = 50.0 grams

Mass CuSO4 = 50.0 grams

Molar mass Zn = 65.38 g/mol

Molar mass CuSO4 = 159.61 g/mol

Step 2: The balanced equation

Zn + CuSO4 → Cu + ZnSO4

Step 3: Calculate moles

Moles = mass / molar mass

Moles Zn = 50.0 grams / 65.38 g/mol = 0.765 moles

Moles CuSO4 = 50.0 grams / 159.61 g/mol = 0.313 moles

Step 4: Calculate limiting reactant

CuSO4 is the limiting reactant. It will completely be consumed (0.313 moles).

Zn is in excess. There will react 0.313 moles

There will remain 0.765 - 0.313 = 0.452 moles

Step 5: Calculate moles products

For 1 mol Zn we need 1 mol CuSO4 to produce 1 mol Cu and 1 mol ZnSO4

For 0.313 moles CuSO4 we'll have 0.313 moles Cu and 0.313 moles ZnSO4

Step 6: Calculate mass products

Mass Cu = 0.313 moles * 63.546 g/mol = 19.89 grams

Mass ZnSO4 = 0.313 moles * 161.47 g/mol = 50.54 grams

3) NH4cl+NaOH -------------->NH3+H2O+NaCl

Step 1: Data given

Mass NH4Cl = 50.0 grams

Mass NaOH = 50.0 grams

Molar mass NH4Cl = 53.49 g/mol

Molar mass NaOH = 40.0 g/mol

Step 2: The balanced equation

NH4Cl + NaOH → NaCl + NH3 + H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles NH4Cl = 50.0 grams / 53.49 g/mol = 0.935 moles

Moles NaOH = 50.0 grams / 40.0 g/mol = 1.25 moles

Step 4: Calculate limiting reactant

NH4Cl is the limiting reactant. It will completely be consumed (0.935 moles).

NaOH is in excess. There will react 0.935 moles

There will remain 1.25 - 0.935 = 0.315 moles

Step 5: Calculate moles products

For 1 mol NH4Cl we need 1 mol NaOH to produce 1 mol NaCl, 1 mol NH3 and 1 mol H2O

For 0.935 moles NH4Cl we'll have 0.935 moles NaCl, 0.935 moles NH3 and 0.935 moles H2O

Step 6: Calculate mass products

Mass NaCl = 0.935 moles * 58.44 g/mol = 54.6 grams

Mass NH3 = 0.935 moles * 17.03 g/mol = 15.9 grams

Mass H2O = 0.935 moles * 18.02 g/mol = 16.8 grams

4) Fe2O3+CO ------------>Fe+CO2

Step 1: Data given

Mass Fe2O3 = 50.0 grams

Mass CO = 50.0 grams

Molar mass Fe2O3 = 159.69 g/mol

Molar mass CO = 28.01 g/mol

Step 2: The balanced equation

Fe2O3 + 3CO → 2Fe + 3CO2

Step 3: Calculate moles

Moles = mass / molar mass

Moles Fe2O3 = 50.0 grams / 159.69 g/mol = 0.313 moles

Moles CO = 50.0 grams / 28.01 g/mol = 1.785 moles

Step 4: Calculate limiting reactant

Fe2O3 is the limiting reactant. It will completely be consumed (0.313 moles).

CO is in excess. There will react 3* 0.313 = 0.939 moles

There will remain 1.785 - 0.939 = 0.846 moles

Step 5: Calculate moles products

For 1 mol Fe2O3 we need 3 moles CO to produce 2 moles Fe, 3 moles CO2

For 0.313 moles Fe2O3 we'll have 0.626 moles Fe and 0.939 moles CO2

Step 6: Calculate mass products

Mass Fe = 0.626 moles * 55.845 g/mol = 35.0 grams

Mass CO2 = 0.939 moles * 44.01 g/mol = 41.3 grams

8 0

3 years ago