in the reaction NO3+CaCl2=2AgCl+Ca(NO3)2, how many grams of excess will remain when 20 grams of AgNO3 are reacted with 15 grams of CaCl2
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To determine the grams of excess that will remain, we need to first find the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
Step 1: Convert the given masses of AgNO3 and CaCl2 to moles.
- AgNO3: 20 grams = (20 g) / (169.87 g/mol) = 0.1177 mol
- CaCl2: 15 grams = (15 g) / (110.98 g/mol) = 0.1351 mol
Step 2: Use the balanced equation to determine the stoichiometric ratio between AgNO3 and CaCl2.
From the balanced equation: 1 mole of AgNO3 reacts with 1 mole of CaCl2
Step 3: Compare the moles of AgNO3 and CaCl2 to identify the limiting reactant.
Since the mole ratio is 1:1, the reactant with the fewer moles is the limiting reactant. In this case, AgNO3 has fewer moles (0.1177 mol), so it is the limiting reactant.
Step 4: Calculate the amount of excess reactant remaining.
To find the moles of CaCl2 that reacted, we can use stoichiometry.
From the balanced equation: 1 mole of AgNO3 reacts with 1 mole of CaCl2, so 0.1177 mol of AgNO3 reacts with 0.1177 mol of CaCl2.
The moles of CaCl2 reacted can be calculated as:
0.1177 mol AgNO3 × (1 mol CaCl2 / 1 mol AgNO3) = 0.1177 mol CaCl2
The excess moles of CaCl2 can be calculated as:
0.1351 mol - 0.1177 mol = 0.0174 mol
Step 5: Convert the excess moles of CaCl2 to grams.
The molar mass of CaCl2 is 110.98 g/mol.
Mass of excess CaCl2 = 0.0174 mol × 110.98 g/mol = 1.93 grams
Therefore, approximately 1.93 grams of excess CaCl2 will remain after the reaction is complete.