Pb(NO3)2 (aq) + 2 NaI (aq) --> PbI2 (s) + 2 NaNO3 (aq)
Starting with with 200.0 grams of Pb(NO3)2 and 120.0 grams of NaI:
A. What is the limiting reagent?
B. How many grams of PbI2 is theoretically formed?
C. How many grams of the excess reactant remains?
D. If 48 grams of NaNO3 actually formed in the reaction, what is the percent yield of this reaction?
Answer:
83.69 gm
Explanation:
molar weight of N2 = 28
Find the number of moles then multiply by this
1.8 x10^24/ (6.022x10^23) * 28 =83.69 gm
Magnesium + Hydrocloric acid -> Magnesium chloride + hydrogen
You can observe a single displacement reaction
"Describe to show that the has formed is hydrogen"
I don't know what you mean. I can show the chemical equation though.
Mg(s) + 2 HCl(aq) --> MgCl 2(aq) + H 2(g)
