CHEMISTRY | MOLE CONCEPT | Question 29
Lead nitrate decomposes on heating as shown in the equation below.
Pb (NO3)2 heat → 2PbO(S) + 4N02(S) + O2(g)
112dm3 of oxygen gas were collected at STP when a sample of lead nitrate was completely decomposed by heating.
Calculate;
a. The mass of the lead nitrate sample.
b. The mass of lead (II) oxide produced.
c. The volume of nitrogen dioxide gas produced at STP.
(Pb = 207, Na =14, 0 = 16; molar volume of a gas at STP = 22.4 dm3)
-
DATA
Volume of O2 = 112dm3
Molar volume at STP = 22.4 dm3
Asked
a. Mass of Pb(NO3)2
b. Mass of PbO
c. Volume of NO2
d. Equation for the reaction
2Pb (NO3)2(s) → 2PbO(s) + 4NO(g) + O2(g)
To convert 112dm3 of O2 into mole
Mole = volume / Molar volume at STP
= 5 mole of O2
-
A. To find mass of Pb(NO3)2 from mole equivalent
2 mole of Pb (NO3)2 Ξ 1 mole O2
?? Ξ 5 moles O2
= (2×5) / 1
= 10 moles of Pb(NO3)2
Mass = mole × molar mass
= 10 moles × 331g/mole
= 3310g
3310g of lead nitrate Pb(NO3)2 was used
More questions
CHEMISTRY questions are much EASIER when you clearly understand the concept. With Smart Darasa exprole concepts in fun and interactive way.
Learn more »Send Your CHEMISTRY Question
Are you a STUDENT | PARENT | TEACHER, Send your CHEMISTRY question for FREE and we will get back to you with the solution or relavant link with similar question. We receive questions every day from 08:00hrs to 16:00 EAT. Depend on volume of questions, we send solutions the same day from 18:00hrs to 20:00hrs EAT or later next day.