Methanol CH3OH is the simplest of the alcohols . It is sinthesized by the reaction of hydrogen and carbon monoxide. CO + H2 > CH3OH

a) If 500 ml CO and 750 mol H2 are present, which is the limiting reactant?

b) How many mols of excess reactant remain unchange?

c) How many moles of CH3OH are formed ?

a) CO is the limiting reactant.

b)

c)

Explanation:

Hello,

In this case, the balanced chemical reaction:

So we proceed as follows:

a) We first compute the moles of CO by using its density (0.00114 g/mL) and molar mass (28 g/mol):

Next, with the given 750 moles of hydrogen, we can compute the moles of carbon monoxide that are consumed by such amount of hydrogen by using their 1:2 molar ratio:

Thus, we see a clear excess of hydrogen, for that reason the carbon monoxide is the limiting reactant.

b) In this case, we first compute the moles of CO that are not consumed:

Next, we use the 1:2 molar ratio again to compute the unchanged moles of hydrogen which is the excess reactant:

c) Finally, we use the reacting moles of carbon monoxide to compute the formed moles of methanol by using the 1:1 molar ratio:

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