Question

Calculate the volume of Hydrogen liberated when 230 gm of sodium reacts with an excess of water ar STP. (Atomic Masses of Sodium = 23 U)
Oxygen = 16U, Hydrogen = 1U
($Na+H_{2O}\to NaOH+H_2)$

Answer 1

Given that,

Atomic masses of sodium $=23u$

Oxygen $=16u$

Hydrogen $=1u$

Now,

$2N{a}_{\left(g\right)}+2H_{2}O\to 2NaO{H}_{\left(aq\right)}+H_2\left(g\right)\uparrow$

$(2\times 23)u+2(2\times 1+1\times 16)u\to 2(23+16+1)u+(2\times 1)u$

$46u+36u\to 80u+2u$

$46g+36g\to 80g+2g$

As per the balanced equation

$46g$ of Na = 2g of Hydrogen

230 g of Na $=230\times \frac{2g}{46}=10g$ of hydrogen

I gram molar mass of any gas at STP

22.4 liters know as gram molar volume

2.0 g of hydrogen occupies $=10.0g\times \frac{22.4}{2.0g}=112$ liters

2 g of hydrogen, 1 mole of $H_2$ contains $6.02\times {10}^{23}$ x H molecules $10g$ of hydrogen contain

$=\frac{10.0g\times 6.02\times {10}^{23}}{2.0g}$

$=30.10\times {10}^{23}$

$=3.01\times {10}^{24}molecules$

Hence, this is the required solution