The balanced chemical reaction:
<span>Cu + 2AgNO3 = Cu(NO3)2 + 2Ag
</span>
We are given the amount of the reactants to be used for the reaction. These values will be the starting point of our calculations.
9.85 g Cu ( 1 mol Cu / 63.55 g Cu ) = 0.15 mol Cu
31.0 g AgNO3 ( 1 mol AgNO3 / 169.87 g AgNO3 ) = 0.18 mol AgNO3
The limiting reactant is AgNO3.
0.18 mol AgNO3 ( 1 mol Cu(NO3)2 / 2 mol AgNO3 ) (187.56 g / 1 mol) =16.88 g Cu(NO3)2
0.15 mol Cu - 0.18 mol AgNO3 ( 1 mol Cu / 2 mol AgNo3) = 0.06 mol Cu excess
<span>0.06 mol Cu ( 63.55 g Cu / 1 mol Cu ) = 3.81 g Cu excess</span>
Answer:
6 x 10 (power to the 15) H2
Answer:
0.375 M
Explanation:
NaOH(aq) + HBr(aq) ------------> NaBr(aq) + H2O(l)
Concetration of acid CA= 0.250M
Concentration of base CB= ????
Volume of acid VA= 30.0mL
Volume of base VB= 20.0mL
Number of moles of acid nA= 1
Number of moles of base nB= 1
CA VA/CB VB= nA/nB
CB= CAVAnB/VB nA
= 0.25× 30×1/20×1= 0.375 M
