Answer:
0.690 liters is the volume of hydrogen gas produced if 2.00 grams of zinc is used with an excess of hydrochloric acid.
Explanation:
Zn(s)+2HCl(aq)\rightarrow ZnCl_2(aq)+H_2(g)Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g)
Mole so zinc = \frac{2.00}{65 g/mol}=0.03077 mol65g/mol2.00=0.03077mol
According to reaction, 1 mole of zinc gives 1 mole of hydrogen gas.
Then 0.03077 mole of zinc will give :
\frac{1}{1}\times 0.03077 mol=0.03077 mol11×0.03077mol=0.03077mol of hydrogen gas
Pressure of hydrogen gas ,P= 1 atm
Temperature of of hydrogen gas ,T= 273.15 K
Volume of hydrogen gas = V = ?
Moles of hydrogen gas = 0.03077 mol
PV = nRT (Ideal gas equation )
V=\frac{nRT}{P}=\frac{0.03077 mol\times 0.0821 atm L/mol K\times 273.15 K}{1 atm}V=PnRT=1atm0.03077mol×0.0821atmL/molK×273.15K
V = 0.690 L
