Unit 1B, Test Review]

siew

1(9.0 % )+2(12.0% 4)+6(1.0%..)+7(16.0g/mm)=151.0 /ad

b) % (=24033x100% = 15,9%

c)3,21 ghydrate, I na kredrate !And Belion 103,0gBEGOU

- 151.0 shydrats" Ino londrests Iand Belson

= 2.19 g Belaou

e

@ 1939 Mag Poux 64.

0 0 =75.390

164,0 g Najpoy :

adecuasto..$.23x1824

CO2 I mol COL . . .

6 382 gCO2 In con 62

o 118g Nassosx(x22 22 molecules con

642 Ss mas 6.022102 atores

2 mas lalo ators

- 138 .2

= 8,98 x1023 atomes

& mass C in iron corpsunda moss a in Asce

:.: =304,8 g Asu 35.5 a 75,5 g

148.4 g Agu

mass Fe =134.8 g total-755gC2= 59.3 g fe

mol - 1

. . . 1.06ms

1.06 mol

- 35.5 sce

.59,3 gFe-Im r=1.06mol Fé .1.06mlEL

75,5 g x lunch or 2.13mince 2.13 moich:

[Fechal

mass.Ba in original compound= mass Ba in Balroy

R = 2.012 ,BaCroup 137.3g Beaca = 1,091, Ba

:b) mass o = 1.345 gtotal-1,011g Ba = .254 , o

c) 1,091, Bax ImeBar = . 007946mlBa .007845 mol Ba

007946 mon .

254 g x me 01588 malo

887946 m .

= 01588 mano...

(BaO)

a) .843 Langairx l.18g dueano 1995 g dire air.

by mass of empty flask =187,20 97.995 g = 156.71

c) mass of unknown gas=158,08g-156.71g=1,37g

di moler mass=1:32 =40,0 %

e) % enor = 140x21-440%/ _=8,9%

44.0 %

- mas 392 mol

in iso l . Ima py

124,0gPy 2mil (as(poudz 310.3 g CagCroudz= 75.1

1mol Pu Iman caglpoud Cazlloudz.

b)1gSiol. ImoSion 10 moico 26.00 = 940.g(0

ul ca - 28.0

:() % . 75.6 g actual

Yield = x100% = 80.4 %

94.0 theorefree!

. . . 601a SiO2 6 molsio Imapy .

.:d) massPy=45.6gSiO2,ImalS:Ur Imolhy 1240spy=15.7gfy

• Massby= 38.2g Cx lm

ncilmarc ilmalPy 124,0 gPy =39.4.gpy

n o

Since the SiO2 formsless. Py, Sioris the limiting

: reactant and a maximum of 15.1, Pe can form,

g. 10mooc mul py ..

SS Co. Ima (oz Imoi CaCO3 loo.islacos,se

- 44.0 gCO2 imalou Iwasralo,

8.99g Glozi

3 x100% =48,1%

3 in sample.- 18.7 g total

: % SiO2 = 100.% -48.1% =51.9 % .:

GOO a) mass H in anilinea mass H in product the

= 6,63g H2o,20 st = 737 gH :

mass N in aniline = mass N in product Ne ...

. . gN . . . . . . . . . .

.:28.0

. = 146 a Nox2 28.0 g N . !

A = 1.46 a N : . : .

mass (= 9.71 9 -1737 g-1.46g=7.51 ..

Mul C = 7.51

. . a C Imola

. 12.

0 6 .626 mol C = 6

- .626 malc. 626 mol c . .

104 more

mal H = 737. H. Imolt

.? lo g.ne = 737 mal H .

I

.

2 1737much .7 :

you mor 1 . CLAN6km

. 104 mol

=1:46 3 Nx 1 mol N = 104mln.

14.0

. . . s W .104 mol N. . .104 mar a

low

: :

Molar. mass .

empirical mass .

n 93 ama

93 ama

mass. O = 750. g Kozx 1molKom 3 moi Onx320 g02

. 71.1 gko, 4 maxko, Imol O2

X 253 g O

:* : The maximum amount of oi that could be produced is more

:: : than what hasbeen produced, so not all of the Koz hasreacted.

# additional O to be produced =253 g =195g = 58.g.02.

mass Ko z reacted=!95gOzxIima ozx mlKor 11.1sko2 578gKoį

320go2 3 mol or mailko.

: massKoż remaining= 750.,Kon:-578gkoz= 172 zKoz

(12 a) No. If the pores are toolange, less solid will be captured

.:. . by the filter paper,which will decrease the mass.

by

. : Yes. Without rousing, the moisture on the solid will contain

dissolved ions. When thewater is removed, these ionswill

reform solid which will increase the mass.

paper.

c) No.

...the If some solid never makes it to the filter

most recorded willbe lower