1 Answers
Call
Talk Doctor Online in Bissoy App
What mass of precipitate (in g) is formed when 58.2 mL of 0.500 M FeCl₃ reacts with excess AgNO₃ in the following chemical reaction? FeCl₃(aq) + 3 AgNO₃(aq) → 3 AgCl(s) + Fe(NO₃)₃(aq)
Mass of precipitate =2.37g
1 Answers 1 views
If 108.5 moles of FeCl3 is combined with 76.8 moles of AgNo3, how many moles of Agcl is produced? With solution
3AgNO3 + 3FeCl3 = 3AgCl + 3Fe(NO3)Mole ratio of FeCl3:AgCl = 1:1Mole ratio of AgNO3:AgCl = 1:1108.5+76.8 = 185.3moles
1 Answers 1 views
Given the following reaction:
Cu+2AgNO3=2Ag+Cu(NO3)2
How many moles of zag will be produced from 46.0 g of Cu, assuming AgNO3 is available in excess?
First, the amount of Cu should be calculated:n = m / Mr,where m - mass, Mr - molecular weight, n - number of moles.n (Cu) = 46 g / 64...
1 Answers 1 views
How many moles of barium metal react to produce 0.100mol of barium nitride? Hpw many moles of nitrogren gas react?
Ba(s) + N2(g) -> Ba3N2(s)
x y 0.100 3Ba(s) + 2N2(g) ------gt; Ba3N2(s) 3 2 1 x=3*0.100/1=0.300 mol y=2*0.100/2-0.200 mol
1 Answers 1 views
How many grams of copper are required to react with 45.0 grams of silver nitrate? Cu+ 2 AgNO3= Cu(NO3)2 + 2 Ag
Solution:The balanced chemical equation:Cu + 2AgNO3= Cu(NO3)2 + 2AgAccording to the equation: n(Cu) = n(AgNO3)/2Moles of AgNO3 = Mass of AgNO3 / Molar mass of AgNO3The molar mass of AgNO3...
1 Answers 1 views
Need the ionic equation and net ionic equations
1. AgNO3(aq) +KCl(aq) ----> KNO3 + AgCl
2. CuCl2(aq) + Ba(NO3)2(aq) -----> Cu(NO3)2 + BaCl2
Thank you so much for helping me!
1. AgNO3(aq) +KCl(aq) ----> KNO3(aq) + AgCl (s) Ag+ + NO3- +K+ + Cl- ----> K+ + NO3- + AgCl(s) Ag+ + Cl- ----> AgCl(s) 2. CuCl2(aq) + Ba(NO3)2(aq) ----->...
1 Answers 1 views
How many moles of lithium would be needed to react with 0.42 mol of Cu(NO3)2
2Li + Cu(NO3)2 = 2LiNO3 + Cun (Li) = 0.84 mole
1 Answers 1 views
How many grams of silver bromide, AgBr, are produced if 15.0 mL of 0.350 M CaBr2(aq) completely react given the reaction:
2 AgNO3(aq) + CaBr2(aq) → Ca(NO3)2(aq) + 2 AgBr(s)
n(CaBr2)= V(CaBr2)×c(CaBr2)=0.015×0.35=0.00525 moln(AgBr)=2×n(CaBr2)=2×0.00525=0.0105 molm(AgBr)=n(AgBr)×M(AgBr)=0.0105×188=1.974 g.
1 Answers 1 views