# Chemistry Reacting masses

I would like to know the answer of the following questions

1. Copper can be obtained from copper(II) oxide and now it is required to produce 508g of copper. 2CuO(s) + C(s) -----------> 2Cu(s) +CO2(g)

a. what mass of copper(II) oxide is required for the production?

b. calculate the mass of carbon dioxide formed in the production.

2. Fe3O4(s) + 4C(s) -------> 3Fe(s) + 4CO(g)

calculate the mass of carbon used to reduce 5000g of the ores if it contains 90% of Fe3O4.

3. Hg(l) + Br2(l) -----------> HgBr2(s)

if 21.5g of mercury (II) bromide produced and the mass of reactant left.

4. N2(g) + 3H2(g) ----------> 2NH3(g)

a. calculate the theoretical yield of ammonia produced from 780g of hydrogen

b, the process is only 15%efficient. what is the mass of ammonia produced?

Update:

thank you for reminding me that

3. Hg(l) + Br2(l) -----------> HgBr2(s)

if 21.5g of mercury (II) react with 15.6g of bromine, calculate the mass of mercury(II) bromide produced and the mass of reactant left.

Rating

1. a.

Molar mass of Cu = 63.5 g mol-1

Molar mass of CuO = 63.5 + 16 = 79.5 g mol-1

Molar mass of CO2 = 12 + 16x2 = 44 g mol-1

2CuO + C → 2Cu + CO2

Mole ratio CuO : Cu : CO2 = 2 : 2 : 1

No. of moles of Cu formed = mass/(molar mass) = 508/63.5 = 8 mol

No. of moles of CuO required = 8 x (2/2) = 8 mol

Mass of CuO required = mol x (molar mass) = 8 x 79.5 = 636 g

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1.b.

No. of moles of Cu formed = 8 mol

No. of moles of CO2 formed = 8 x (1/2) = 4 mol

Mass of CO2 formed = mol x (molar mass) = 4 x 44 = 176 g

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2.

Molar mass of C = 12 g mol-1

Molar mass of Fe3O4 = 55.9x3 + 16x4 = 231.7 g mol-1

Fe3O4 + 4C → 3Fe + 4CO

Molar ratio Fe3O4 : C = 1 : 4

Mass of Fe3O4 = 5000 x 90% = 4500 g

No. of moles of Fe3O4 used = mass/(molar mass) = 4500/231.7 = 19.42 mol

No. of moles of C used = 19.42 x 4 = 77.68 mol

Mass of C used = mol x (molar mass) = 77.68 x 44 = 3418 g

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3.

Molar mass of Hg = 201 g mol-1

Molar mass of Br2 = 80 x 2 = 160 g mol-1

Molar mass of HgBr2 = 201 + 80x2 = 361 g mol-1

Hg+ Br2 → HgBr2

mole ratio Hg : Br2 : HgBr2 = 1 : 1 : 1

No. of moles of Hg added = mass/(molar mass) = 21.5/201 = 0.1070 mol

No. of moles of Br2 added = mass/(molar mass) = 15.6/160 = 0.0975 mol

Br2 is the limiting reactant, and Hg is in excess.

No. of moles of Br2 reacted = 0.0975 mol

No. of moles HgBr2 formed = 0.0975 mol

Mass of HgBr2 formed = mol x (molar mass) = 0.0975 x 361 = 35.2 g

No. of moles of Br2 reacted = 0.0975 mol

No. of moles of Hg reacted = 0.0975 mol

Mass of Hg reacted = mol x (molar mass) = 0.0975 x 201 = 19.6 g

Mass of Hg left = 21.5 - 19.6 = 1.9 g

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4.a.

Molar mass of H2 = 1x2 = 2 g mol-1

Molar mass of NH3 = 14 + 1x3 = 17 g mol-1

N2 + 3H2 → 2NH3

Mole ratio H2 : NH3 = 3 : 2 (assuming 100% yield)

No. of moles of H2 added = mass/(molar mass) = 780/2 = 390 mol

No. of moles of NH3 formed if 100% yield = 390 x (2/3) = 260 mol

Theoretical yield of NH3 = mol x (molar mass) = 260 x 17 = 4420 g

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4.b.

Mass of NH3 produced = 4420 x 15% = 663 g