Answer:
- 13.56 g of sodium chloride are theoretically yielded.
- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.
- 0.50 g of sodium nitrate remain when the reaction stops.
- 92.9 % is the percent yield.
Explanation:
Hello!
In this case, according to the question, it is possible to set up the following chemical reaction:
Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:
Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:
Therefore, the leftover of sodium nitrate is:
Finally, the percent yield is computed via:
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Answer:
D. all of these are chemical changes
Answer:
V = 85.619 L
Explanation:
To solve, we can use the ideal gas law equation, PV = nRT.
P = pressure (645 mmHg)
V = volume (?)
n = amount of substance (3.00 mol)
R = ideal gas constant (62.4 L mmHg/mole K)
T = temperature (295K)
Now we would plug in the appropriate numbers into the equation using the information given and solve for V.
(645)(V) = (3.00)(62.4)(295)
(V) = (3.00)(62.4)(295)/645
V = 85.619 L
Answer:
100. mL
Explanation:
Step 1: Write the balanced equation for the double displacement reaction
CaCl₂ + Na₂CO₃ ⇒ 2 NaCl + CaCO₃
Step 2: Calculate the moles corresponding to 1.00 g of CaCO₃
The molar mass of CaCO₃ is 100.09 g/mol.
1.00 g × 1 mol/100.09 g = 0.0100 mol
Step 3: Calculate the moles of CaCl₂ required to produce 0.0100 moles of CaCO₃
The molar ratio of CaCl₂ to CaCO₃ is 1:1. The moles of CaCl₂ required are 1/1 × 0.0100 mol = 0.0100 mol.
Step 4: Calculate the volume of 0.100 M CaCl₂ that contains 0.0100 mol
0.0100 mol × 1 L/0.100 mol × 1000 mL/1 L = 100. mL
