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3 years ago

How many dm³ of hydrogen,measured at s.t.p.,would be needed to reduce 47.7g of copper(II) oxide to copper?

Chemistry

1 answer:

GaryK [48]3 years ago

4 0

Answer:

Option D. 13.44

Explanation:

We'll begin by calculating the number of mole in 47.7g of copper(II) oxide, CuO.

This can be obtained as follow:

Mass of CuO = 47.7 g

Molar mass of CuO = 63.5 + 16 = 79.5 g/mol

Mole of CuO =.?

Mole = mass /Molar mass

Mole of CuO = 47.7/79.5

Mole of CuO = 0.6 mole

Next, we shall write the balanced equation for the reaction. This is given below:

CuO + H2 —> Cu + H2O

From the balanced equation above,

1 mole of CuO reacted with 1 mole of H2 to produce 1 mole of Cu and 1 mole of H2O.

Next, we shall determine the number of mole of H2 needed to react completely with 0.6 mole of CuO.

This can be obtained as follow:

From the balanced equation above,

1 mole of CuO reacted with 1 mole of H2.

Therefore, 0.6 mole of CuO will also react with 0.6 mole of H2.

Finally, we shall determine the volume occupied by 0.6 mole of H2 at STP.

This can be obtained as follow:

1 mole of H2 occupied 22.4 dm³ at STP.

Therefore, 0.6 mole of H2 will occupy = 0.6 × 22.4 = 13.44 dm³.

Therefore, 13.44 dm³ of H2 is needed for the reaction.

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3 years ago

Determine the empirical formulas for compounds with the following percent compositions:

Elis [28]

Answer:

For a: The empirical formula of the compound is $P_2O_5$

For b: The empirical formula of the compound is $KH_2PO_4$

Explanation:

For a:

We are given:

Percentage of P = 43.6 %

Percentage of O = 56.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of P = 43.6 g

Mass of O = 56.4 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Phosphorus = $\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{43.6g}{31g/mole}=1.406moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{56.4g}{16g/mole}=3.525moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.406 moles.

For Phosphorus = $\frac{1.406}{1.406}=1$

For Oxygen = $\frac{3.525}{1.406}=2.5$

Converting the moles in whole number ratio by multiplying it by '2', we get:

For Phosphorus = $1\times 2=2$

For Oxygen = $2.5\times 2=5$

Step 3: Taking the mole ratio as their subscripts.

The ratio of P : O = 2 : 5

Hence, the empirical formula for the given compound is $P_2O_5$

For b:

We are given:

Percentage of K = 28.7 %

Percentage of H = 1.5 %

Percentage of P = 22.8 %

Percentage of O = 56.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of K = 28.7 g

Mass of H = 1.5 g

Mass of P = 43.6 g

Mass of O = 56.4 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Potassium = $\frac{\text{Given mass of Potassium}}{\text{Molar mass of Potassium}}=\frac{28.7g}{39g/mole}=0.736moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of hydrogen}}=\frac{1.5g}{1g/mole}=1.5moles$

Moles of Phosphorus = $\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{22.8g}{31g/mole}=0.735moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{47g}{16g/mole}=2.9375moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.735 moles.

For Potassium = $\frac{0.736}{0.735}=1$

For Hydrogen = $\frac{1.5}{0.735}=2.04\approx 2$

For Phosphorus = $\frac{0.735}{0.735}=1$

For Oxygen = $\frac{2.9375}{0.735}=3.99\approx 4$

Step 3: Taking the mole ratio as their subscripts.

The ratio of K : H : P : O = 1 : 2 : 1 : 4

Hence, the empirical formula for the given compound is $KH_2PO_4$

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3 years ago