Question

Solid sodium reacts violently with water, producing heat, hydrogen gas, and sodium hydroxide. How many molecules of hydrogen gas are formed when 48.7 g of sodium are added to water? Show your work. (4 points) 2Na + 2H2O ---> 2NaOH + H2 2Na + 2H2O ---> 2NaOH + H2 48.7 g ? Molecules 2 mol Na= 1 mol H2 1 mol Na= 23g Na 1 mol H2= 6.02x1023 molecules (48.7g Na/1)x(1 mol Na/23g Na)x(1 mol H2/2mol Na)x(6.02x1023/1 mol H2) (48.7)(6.02x1023)/(23)(2)= 293.174x1023 molecules Can someone please tell me what I did wrong?

Answer 1

Answer: Number of molecules of hydrogen gas $6.32\times 10^{32}$

Explanation:

$2Na+2H_2O\rightarrow 2NaOH+H_2$

Number of moles of sodium =$\frac{\text{mass of sodium}}{\text{molar mass of sodium}}=\frac{48.7 g}{23 g/mol}=2.11mol$

According to reaction , 2 moles of sodium produces 1 mole of hydrogen gas , then 2.11 mol of sodium will= $\frac{1}{2}\times 2.11 mol$ of hydrogen gas that is 1.05 moles of hydrogen gas.

Number of molecules = $N_A(\text{Avogadro number})\times$ moles of substance

Moles of hydrogen gas formed = 1.05 moles

Number of molecules of hydrogen gas = $N_A\times$ moles of hydrogen gas

Number of molecules of hydrogen gas $=6.022\times 10^{23} mole^{-1}\times 1.05 mole=6.32\times 10^{32}$