Respuesta :
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:
[tex]Zn(s)+2HCl(aq)\rightarrow ZnCl_2(aq)+H_2(g)[/tex]
Mole so zinc = [tex]\frac{2.00}{65 g/mol}=0.03077 mol[/tex]
According to reaction, 1 mole of zinc gives 1 mole of hydrogen gas.
Then 0.03077 mole of zinc will give :
[tex]\frac{1}{1}\times 0.03077 mol=0.03077 mol[/tex] 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 )
[tex]V=\frac{nRT}{P}=\frac{0.03077 mol\times 0.0821 atm L/mol K\times 273.15 K}{1 atm}[/tex]
V = 0.690 L
0.690 liters is the volume of hydrogen gas produced if 2.00 grams of zinc is used with an excess of hydrochloric acid.
Answer: 0.685 L
Explanation:
Since the reaction is said to take place at STP, the molar volume 22.4Lmol can be used. First find the number of moles of zinc. Use the periodic table above to determine the formula mass of zinc. The formula mass is 65.38gmol.
2.00gZn × 1.00mol Zn / 65.38g Zn = 0.03059 mol Zn
Convert moles of zinc to moles of hydrogen gas, using the mole ratio from the balanced equation.
0.03059 mol Zn × (1.00 mol H / 21.00 mol Zn) = 0.03059 mol H2
The reaction takes place at STP, so each mole of H2 will occupy a volume of 22.4L. The volume of hydrogen gas produced, rounded to three significant figures, is obtained by the equation below.
0.03059 mol H2 × 22.4 L/mol = 0.685L
