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Step2247 [10]
3 years ago
10

For the following reaction, calculate how many moles of each product are formed when 0.356 moles of PbS completely react. Assume

there is an excess of oxygen.
2PbSs+3O2g→2PbOs+2SO2g
Chemistry
2 answers:
VMariaS [17]3 years ago
7 0

Answer:

0.356 moles of each products (PbO and SO2) are formed.

Explanation:

First step is to write out the equation and ensure it is balanced before proceeding.

2PbSs + 3O2g → 2PbOs + 2SO2g

2            3            2              2

The equation above is balanced because there are equal number of atoms of elemenents on both sides of the arrow.

By stating that oxygen is in excess, this tells that PbS is the limiting reactant (i.e the amount of products formed depends on it) and not on oxygen.

From the equation;

The products formed are PbO and SO2.

For PbO, it requires 2 moles of PbS to react completely in order to form 2 moles of PbO.

For SO2, it requres 2 moles of PbS to react completely in order to form 2 moles of SO2.

This is gotten from the equation.

PbO

2 moles = 2 moles

0.356 moles = x

upon cross multiplication, we are left with;

x = (0.356 * 2) / 2 = 0.356 moles

SO2

2 moles = 2 moles

0.356 moles = x

upon cross multiplication, we are left with;

x = (0.356 * 2) / 2 = 0.356 moles

galben [10]3 years ago
3 0

Answer:

When 0.356 moles of PbS completely react 0.356 moles of PbO and 0.356 moles of SO₂ are produced.

Explanation:

2PbS (s) + 3O₂ (g) → 2PbO (s) + 2SO₂ (g)

2 mol PbS ____________ 2 mol PbO

0.356 mol PbS ________         x

          x = 0.356 mol PbO

2 mol PbS ____________ 2 mol SO₂

0.356 mol PbS ________         x

          x = 0.356 mol SO₂

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CaBr + KOH – Ca(OH), + KBr (balance first) What mass, in grams, of
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Answer:

129.73 g of CaBr₂

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

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Next, we shall determine the mass of CaBr₂ that reacted and the mass of Ca(OH)₂ produced from the balanced equation. This can be obtained as follow:

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Mass of CaBr₂ from the balanced equation = 1 × 200 = 200 g

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From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Finally, we shall determine the mass of CaBr₂ that react when 48 g of Ca(OH)₂ were produced. This can be obtained as follow:

From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

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