Question

Q.1 If 3.5 Moles of Ca(OH)^2 are present in the reaction mixture how many moles of sulfuric acid are needed to complete the equation?
Q.2 What will happen to Mg O reacted with copper?Explain your answer.
Q.3 Which compound has 48 Gm of carbon and 12 Gm of Hydrogen if the R M M is 30?
Q.4 When sodium carbonate is added to magnesium chloride solution the reaction that occur is represented by which equation?
Q.5 Which compound has 3 Gm of carbon and 0.5 Gm of Hydrogen if R M M is 42?
Q.6 20 Gm is burned.what mass of sodium oxide is formed?

Answer 1

Answer on Question #44707–Chemistry–Organic Chemistry

Questions

Q.1

If 3.5 moles of Ca(OH)₂ are present in the reaction mixture how many moles of sulfuric acid are needed to complete the equation?

Q.2

What will happen to MgO reacted with copper? Explain your answer.

Q.3

Which compound has 48 g of carbon and 12 g of hydrogen if the RMM is 30?

Q.4

When sodium carbonate is added to magnesium chloride solution the reaction that occur is represented by which equation?

Q.5

Which compound has 3 g of carbon and 0.5 g of hydrogen if RMM is 42?

Q.6

20 g [of sodium] is burned. What mass of sodium oxide is formed?

Answers

Q.1

Ca(OH)₂ + H₂SO₄ → CaSO₄ + 2 H₂O

As is clear from the chemical equation Ca(OH)₂ and H₂SO₄ react in equimolar ratio. So, 3.5 moles of sulfuric acid are needed for 3.5 moles of Ca(OH)₂ to react completely.

Q.2

No reaction occurs, because Mg is more active (it is oxidized more easily) than Cu. And red-ox reaction would occur in reverse direction only

$$\begin{array}{l} \mathrm{CuO} + \mathrm{Mg} \rightarrow \mathrm{Cu} + \mathrm{MgO} \\ \mathrm{MgO} + \mathrm{Cu} \rightarrow \text{no reaction} \end{array}$$

Q.3

$$\begin{array}{l} \mathrm{M(C_xH_y)} = 12 \cdot x + 1 \cdot y = 30 \text{ g/mol} \\ n(\mathrm{C}) = m(\mathrm{C}) / \mathrm{M}(\mathrm{C}) = 48 \text{ g} / 12 \text{ g/mol} = 4 \text{ mol} \\ n(\mathrm{H}) = m(\mathrm{H}) / \mathrm{M}(\mathrm{H}) = 12 \text{ g} / 1 \text{ g/mol} = 12 \text{ mol} \\ x / y = n(\mathrm{C}) / n(\mathrm{H}) = 4 / 12 = 1 / 3, \text{ whence } y = 3x. \end{array}$$

Thus, we have the set of two linear equations:

$$\left\{ \begin{array}{l} 12x + y = 30 \\ y = 3x \end{array} \right.$$ $$\begin{array}{l} 12x + 3x = 30 \\ 15x = 30 \\ x = 30 / 15 = 2 \\ y = 3 \cdot 2 = 6 \end{array}$$

So, the compound is C₂H₆ (ethane)

Q.4

$$\mathrm{Na_2CO_3} + \mathrm{MgCl_2} \rightarrow 2 \mathrm{NaCl} + \mathrm{MgCO_3} \text{ (s)}$$

Q.5

$$\begin{array}{l} \mathrm{M(C_xH_y)} = 12 \cdot x + 1 \cdot y = 42 \ \mathrm{g/mol} \\ \mathrm{n(C)} = \mathrm{m(C)/M(C)} = 3 \ \mathrm{g} / 12 \ \mathrm{g/mol} = 0.25 \ \mathrm{mol} \\ \mathrm{n(H)} = \mathrm{m(H)/M(H)} = 0.5 \ \mathrm{g} / 1 \ \mathrm{g/mol} = 0.5 \ \mathrm{mol} \\ x / y = \mathrm{n(C)/n(H)} = 0.25 / 0.5 = 1 / 2, \text{ whence } y = 2x. \end{array}$$

Thus, we have the set of two linear equations:

$$\left\{ \begin{array}{l} 12x + y = 42 \\ y = 2x \end{array} \right.$$ $$\begin{array}{l} 12x + 2x = 42 \\ 14x = 42 \\ x = 42/14 = 3 \\ y = 2 \cdot 3 = 6 \end{array}$$

So, the compound is $\mathbf{C_3H_6}$ (ethene)

Q.6

$$\begin{array}{l} 20 \ \mathrm{g} \quad \mathrm{m}, \ \mathrm{g} \\ 4 \ \mathrm{Na} + \mathrm{O_2} \rightarrow 2 \ \mathrm{Na_2O} \\ 4 \cdot 23 \ \mathrm{g/mol} \quad 2 \cdot 62 \ \mathrm{g/mol} \\ \mathrm{m(Na_2O)} = \mathrm{m(Na)} \cdot 2 \cdot \mathrm{M(Na_2O)} / 4 \cdot \mathrm{M(Na)} = 20 \cdot 2 \cdot 62 / 4 \cdot 23 = 27 \ \mathrm{g} \end{array}$$

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