Question

a)The number of moles and the mass (in g) of calcium peroxide, CaO₂, needed to produce 2.130 kg of calcium oxide, CaO. (O₂ is the other product.) (Give balanced Equation and order of steps)b)The number of moles and the mass (in g) of C₂H₄ required to react with H₂O to produce 9.51 g of C₂H₆O.(Give balanced Equation and order of steps) c)The number of moles and the mass (in g) of potassium nitrate, KNO₃, required to produce 136 g of oxygen. (KNO₂ is the other product.) (Give balanced Equation and order of steps d)The number of moles and the mass (in g) of carbon dioxide formed by the combustion of 32.0 kg of carbon in an excess of oxygen. (Give balanced Equation and order of steps(e)The number of moles and the mass (in g) of copper(II) carbonate needed to produce 1.200 kg of copper(II) oxide. (CO₂ is the other product.)(f)The number of moles and the mass (in g) of C₂H₄I₂ formed by the reaction of 12.05 g of C₂H₄ with an excess of I₂.

Answer 1

a)

Balanced equation for production of calcium oxide from calcium peroxide:

2CaO₂ → 2CaO + O₂

It is clear that 2moles of calcium peroxide (2x72g) produces 2 moles of calcium oxide (2x56g).

Number of moles of calcium peroxide needed to produce 2.13 kg of calcium oxide will be:

(2molesx2130g)/112g = 38 moles

Mass of calcium peroxide needed to produce 2.13 kg of calcium oxide will be:

(144gx2130g)/112g =2738.5 g = 2.738 kg

b)

Balanced equation will be:

C₂H₄ + H₂O → C₂H₆O

1 mole C₂H₄ (28g) produces 1 mole C₂H₆O (46g)

Number of moles of C₂H₄ needed to produce 9.51 g of C₂H₆O will be:

(9.51g x 1mole)/46 g = 0.206 mole

Mass of C₂H₄ needed to produce 9.51 g of C₂H₆O will be:

(9.51g x 28g)/46 g = 5.788 g

c)

Balanced equation will be:

2KNO₃ → 2KNO₂ + O₂

2 moles of KNO₃(2x101g) produces 1 mole of O₂(32g).

Number of moles of KNO₃ needed to produce 136 g of oxygen will be:

(2molesx136g)/32g = 8.5 mole

Mass of KNO₃ needed to produce 136 g of oxygen will be:

(2x101gx136g)/32g = 858.5 g

d)

Balanced equation will be:

C + O₂ → CO₂

1 mole of CO₂(44g) is formed by the combustion of 1 mole of carbon(12g).

Number of mole of carbondioxide formed by the combustion of 32 kg of carbon will be:

(1molex3200g)/12g = 2666.6 moles

Mass of carbondioxide formed by the combustion of 32 kg of carbon will be:

(44gx3200g)/12g = 117.33 kg

(e)

Balanced equation will be:

CuCO₃ → CuO + CO₂

1 mole of copper carbonate(123.5g) is needed to produce 1 mole of copper oxide(79.5g).

Number of moles of copper carbonate needed to produce 1.200 kg of copper oxide will be:

(1200g x 1mole)/79.5g = 15.09 mole

Mass of copper carbonate needed to produce 1.200 kg of copper oxide will be:

(1200g x 123.5g)/79.5g = 1.864 kg

(f)

Balanced equation will be:

C₂H₄ + I₂ → C₂H₄I₂

1 mole of C₂H₄(28g) produces 1 mole of C₂H₄I₂(280g).

Number of moles of C₂H₄I₂ formed by the reaction of 12.05 g of C₂H₄ will be:

(12.05gx1mole)/28g = 0.4303 mole

Mass of C₂H₄I₂ formed by the reaction of 12.05 g of C₂H₄ will be:

(12.05gx280g)/28g = 120.48 g