The thermite reaction occurs when iron(III) oxide reacts with solid aluminum. The reaction is so hot that molten iron forms as a product.
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Fe2O3(s) + Al(s) → Fe(C) + Al2O3(s)
What mass of aluminum should be used in order to completely
consume 10.0 g Fe2O3(s)? If the reaction described produces 5.3 g
Al2O3(s), what is the percent yield?

Explanation:

Hello.

In this case, for the given balanced reaction:

For 10.0 g of iron (III) oxide (molar mass = 160 g/mol), based on the 1:2 mole ratio with Al (atomic mass = 27 g/mol), the required mass is then:

Moreover, as 5.3 g of aluminum oxide are actually yielded, from the 10.0 g of iron (III) oxide, we can compute the theoretical mass of aluminum oxide (molar mass = 102 g/mol) via their 1:1 mole ratio:

Thus, the percent yield (actual/theoretical*100%) turns out:

Best regards.

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