Answer: 15.33 grams of the excess reactant KBr will remain.
Explanation: The molar mass of Pb(NO3)2 is 331.21 g/mol:
32.5 g / 331.21 g/mol = 0.098 mol Pb(NO3)2
The molar mass of KBr is 119.00 g/mol:
38.75 g / 119.00 g/mol = 0.325 mol KBr
Agreeing to the adjusted condition, 1 mole of Pb(NO3)2 responds with 2 moles of KBr to create 1 mole of PbBr2:
1 mol Pb(NO3)2 : 2 mol KBr : 1 mol PbBr2
In this manner, the restricting reactant is Pb(NO3)2, because it will be totally devoured within the response.
The number of moles of PbBr2 delivered can be calculated utilizing the mole proportion from the adjusted condition:
0.098 mol Pb(NO3)2 x (1 mol PbBr2 / 1 mol Pb(NO3)2) = 0.098 mol PbBr2
The mass of PbBr2 delivered can be calculated utilizing its molar mass of 367.01 g/mol:
0.098 mol PbBr2 x 367.01 g/mol = 35.93 g PbBr2
Subsequently, 35.93 grams of the accelerate PbBr2 will be delivered.
To determine the mass of the abundance reactant, we are able utilize the sum of constraining reactant devoured within the response to find the sum of overabundance reactant remaining.
From the calculation above, we know that 0.098 mol of Pb(NO3)2 was expended within the response. Utilizing the mole proportion from the adjusted condition, we are able calculate the number of moles of KBr required to respond with this sum of Pb(NO3)2:
0.098 mol Pb(NO3)2 x (2 mol KBr / 1 mol Pb(NO3)2) = 0.196 mol KBr
Hence, 0.196 moles of KBr were required to respond with the 0.098 moles of Pb(NO3)2, clearing out an overabundance of KBr:
0.325 mol KBr - 0.196 mol KBr = 0.129 mol KBr remaining
The mass of the overabundance KBr can be calculated utilizing its molar mass of 119.00 g/mol:
0.129 mol KBr x 119.00 g/mol = 15.33 g KBr