How much volume, mass and number of molecules of hydrogen liberated when 230 g of sodium reacts with excess of water at STP. (atomic masses of Na = 23U, O = 16U, and H =1U)

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112 litres, 10 g, $3.01 \times 10^{24}$ molecules

224 litres, 20 g, $3.01 \times 10^{26}$ molecules

56 litres, 10 g, $3.01 \times 10^{24}$

112 litres, 100 g, $3.01 \times 10^{24}$ molecules

answer is A.

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Detailed Solution

The hydration reaction of sodium metal is

$2 Na (s) + 2 H_{2} O (l) \to NaOH (aq) + H_{2} (g)$

At S.T.P one mol of any gas will occupy 22.4 L.

$2 \times 23 g of Na \to 22.4 {L of H}_{2} gas$

$230 g of Na \to' x' {L of H}_{2} gas$

$∴ x = \frac{230 g \times 22.4 L}{2 \times 23 g} = 112 {L of H}_{2}$

We know that

$22.4 {Lof H}_{2} at S.T.P \to 2 {g of H}_{2}$

$112 {L of H}_{2} \to x {g of H}_{2}$

$∴ x = \frac{2 {g of H}_{2}}{22.4 L} \times 112 L = 10.0 {g of H}_{2}$

$2 {g of H}_{2} \to 6.02 \times 10^{23} H_{2} molecules$

$10 {g of H}_{2} \to' x' {of H}_{2} molecules$

$∴ x = \frac{6.02 \times 10^{23} molecules}{2.0 g} \times 10 g = 3.01 \times 10^{24} molecules$

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