calculate how many grams of the products listed would be produced by complete reaction of 0.160 mol of the first reactant. AgNO3(aq) + LiOH(aq) → AgOH(s) + LiNO3(aq)
AgOH
To calculate the number of grams of AgOH produced by the reaction, we need to consider the balanced chemical equation and use stoichiometry.
The balanced chemical equation is:
AgNO3(aq) + LiOH(aq) → AgOH(s) + LiNO3(aq)
From the equation, we can see that the molar ratio between AgNO3 and AgOH is 1:1. This means that for every 1 mole of AgNO3, 1 mole of AgOH is produced.
Given that we have 0.160 mol of AgNO3, we can use this information to calculate the number of moles of AgOH produced:
0.160 mol AgNO3 x (1 mol AgOH / 1 mol AgNO3) = 0.160 mol AgOH
Since we have the number of moles of AgOH, we can now calculate the grams of AgOH using its molar mass.
The molar mass of AgOH is the sum of the atomic masses of its constituent elements:
Ag: 107.87 g/mol
O: 16.00 g/mol
H: 1.01 g/mol
Molar mass of AgOH = (107.87 g/mol) + (16.00 g/mol) + (1.01 g/mol) = 124.88 g/mol
Finally, we can calculate the mass of AgOH produced:
0.160 mol AgOH x (124.88 g AgOH / 1 mol AgOH) = 19.98 g AgOH
Therefore, the complete reaction of 0.160 mol of AgNO3 would produce 19.98 grams of AgOH.