Stoichiometry Interactive 2/20/25

2 Al₂O₃ → 4 Al + 3O₂

Setup the equation to solve for mass of Al₂O₃ from 96.5 g Al.

j

g

h

i

f

e

d

27 g Al

1 mol Al2O3

22.4 L

1 mol Al

32 g O2

4 mol Al

3 mol O2

102 g Al2O3

2 mol Al2O3

96.5 g Al

Tags

112.43.c.9.C

​ In stoichiometry setup, what goes in the numerator of box 1?

• Given #, unit and compound

• 1 mol of given

• 1 mole of final

• Coefficient of given

• coefficient of final

Tags

NGSS.HS-PS1-7

​ In stoichiometry setup, what goes in the numerator of box 4?

• Mole conversion of given unit

• Mole conversion of final unit

• 1 mole of final

• Coefficient of given

• coefficient of final

Tags

NGSS.HS-PS1-7

​ In stoichiometry setup, what goes in the denominator of box 4 (always and forever)?

• Mole conversion of given unit

• Mole conversion of final unit

• 1 mole of final

• 1 mole of given

• coefficient of final

​ In stoichiometry setup, what goes in the numerator of box 3?

• Given #, unit and compound

• Mole conversion of final unit.

• 1 mole of final

• Coefficient of given (from balanced chemical equation)

• coefficient of final (from balanced chemical equation)

Tags

NGSS.HS-PS1-7

​ In stoichiometry setup, what goes in the denominator of box 3?

• Given #, unit and compound

• Mole conversion of given unit.

• 1 mole of final

• Coefficient of given (from balanced chemical equation)

• coefficient of final (from balanced chemical equation)

Tags

NGSS.HS-PS1-7

Put the correct labels in the correct spots.

c

d

b

a

given of Al(OH)3

molar mass of Al(OH)3

molar mass of sodium

mole ratio

Tags

NGSS.HS-PS1-7

Put the correct labels in the correct spots.

c

d

b

a

molar mass of sodium

molar mass of Al(OH)3

mole ratio

given of Al(OH)3

Tags

NGSS.HS-PS1-7

(0901) Which conversion would you use if you need to convert from MOLES to particles?

• $\frac{1\ mole}{6.02\ x\ 10^{23}particles}$

• $\frac{6.02\ x\ 10^{23}moles}{1\ particle}$

• $\frac{6.02\ x\ 10^{23}particles}{1\ mole}$

• $\frac{1\ particle}{6.02\ x\ 10^{23}\ moles}$

(0903) Suzie wants to know the mass of 1.55 moles of nitrogen trichloride. How do you set up the problem?

• $1.55\ mole\ NCl_3\ \times\ \frac{1\ mole}{molar\ mass\ of\ NCl_3\ \left(g\right)}=$

• $1.55\ mole\ NCl_3\ \times\ \frac{molar\ mass\ of\ NCl_3\ \left(g\right)}{1\ mole}=$

• $\ 1\ mole\ \times\ \frac{1.55\ mole\ NCl_3}{molar\ mass\ of\ NCl_3\left(g\right)}=$

• $\ molar\ mass\ of\ NCl_3\left(g\right)\ \times\ \frac{1\ mole}{1.55\ mole\ NCl_3}=$

Tags

NGSS.HS-PS1-7