4 Al + 3 O₂ --> 2 Al2Ol₃ Given 6 moles of aluminum, how many grams of aluminum oxide will be produced?

305.88 g

Explanation:

4 Al + 3 O₂ → 2 Al₂O₃

First, we convert 6 moles of aluminum (Al) into moles of aluminum oxide (Al₂O₃), using the stoichiometric coefficients:

6 mol Al * = 3 mol Al₂O₃

Finally we convert 3 moles of aluminum oxide into grams, using its molar mass:

3 mol Al₂O₃ * 101.96 g/mol = 305.88 g

away from the curb

explanation:

the wheels would then hit the curb if it started to rolling, decreasing the chance of an accident.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol