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:
1, 1, 2, 3
Explanation:
The numbers 1 and 8 both have 1 sig. fig.
The number 13 has 2 sig. figs.
The number 104 has 3 sig. figs.
Answer:
526g is the mass of this sample
Explanation:
To solve this question we must, as first, find the <em>molar mass </em>of Al₂(Cr₂O₇)₃ using the periodic table. The molar mass is defined as the mass of this compound per mole. With this value we can find the mass in 0.750 moles as follows:
<em>Molar mass Al₂(Cr₂O₇)₃</em>
2Al = 2*26.98g/mol = 53.96g/mol
6 Cr = 6*51.9961g/mol = 311.9766g/mol
21 O = 21*15.999g/mol = 335.979g/mol
53.96g/mol + 311.9766g/mol + 335.979g/mol
= 701.9156g/mol
The mass of 0.750 moles is:
0.750 moles * (701.9156g / mol) =
<h3>526g is the mass of this sample</h3>
Answer:
the correct awnser is b 1-heptene