Question

W e applicable 1) Balance the following reaction. _C,H,O + Oai -_CO + H2O a) What mass of O, would be required to react fully with 12.6 g C,H,O?! b) If 5.3 g C,H,O is reacted with 8.85 g 0, identify the limiting reactant. 2) During the lab this week, Magnesium will be reacted with the two acids HCl or H, SO, to produce hydrogen gas and magnesium chloride or magnesium sulfate, respectively. Write out the balanced chemical equations for the two reactions. 3) During the lab this week, a constant amount of Mg metal (-0.080 g) will be reacted with the two acids HCl or H, SO, to produce hydrogen gas. a. Write out a correctly formatted hypothesis that describes how the amount of hydrogen gas produced with be affected by the concentration of acid. b. Write out a correctly formatted hypothesis that describes how the amount of hydrogen gas produced with be affected by the type of acid, HCI or H,SO reacted with the Mg metal.

Answer 1

1)

C4H10O(l) + 6 O2(g) -> 4 CO2(g) + 5 H2O(g)

a)

Molar mass of C4H10O,

MM = 4*MM(C) + 10*MM(H) + 1*MM(O)

= 4*12.01 + 10*1.008 + 1*16.0

= 74.12 g/mol

mass of C4H10O = 12.6 g

mol of C4H10O = (mass)/(molar mass)

= 12.6/74.12

= 0.17 mol

According to balanced equation

mol of O2 required = (6/1)* moles of C4H10O

= (6/1)*0.17

= 1.02 mol

Molar mass of O2 = 32 g/mol

mass of O2 = number of mol * molar mass

= 1.02*32

= 32.64 g

Answer: 32.6 g

b)

Molar mass of C4H10O,

MM = 4*MM(C) + 10*MM(H) + 1*MM(O)

= 4*12.01 + 10*1.008 + 1*16.0

= 74.12 g/mol

mass(C4H10O)= 5.3 g

use:

number of mol of C4H10O,

n = mass of C4H10O/molar mass of C4H10O

=(5.3 g)/(74.12 g/mol)

= 7.151*10^-2 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 8.85 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(8.85 g)/(32 g/mol)

= 0.2766 mol

1 mol of C4H10O reacts with 6 mol of O2

for 7.151*10^-2 mol of C4H10O, 0.429 mol of O2 is required

But we have 0.2766 mol of O2

so, O2 is limiting reagent

Answer: O2

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