Zn(s) + 2HCl(l) ---> ZnCl2(s) + H2(g) at STP
What mass of zinc is needed to produce 115 mL of hydrogen?
0.115L/22.4L/mol = 0.0051 moles H2 gas
0.0051 mol x 65g/mol = 0.33 grams
correct, at stp.
To find the mass of zinc needed to produce 115 mL of hydrogen gas, we first need to determine the number of moles of hydrogen gas using the ideal gas law equation: PV = nRT.
In this case, the temperature and pressure are at Standard Temperature and Pressure (STP). STP is defined as a temperature of 0 degrees Celsius or 273 Kelvin, and a pressure of 1 atm.
Since we have 115 mL of hydrogen gas, we need to convert it to liters by dividing by 1000:
115 mL / 1000 mL/L = 0.115 L
Next, we need to use the molar volume of a gas at STP, which is approximately 22.4 L/mol.
0.115 L / 22.4 L/mol = 0.0051 mol H2 gas
Now, to find the mass of zinc needed, we need to use the stoichiometric coefficients from the balanced chemical equation. The ratio of moles of zinc to moles of hydrogen gas is 1:1. This means that for every 1 mole of hydrogen gas produced, 1 mole of zinc is consumed.
Therefore, the mass of zinc needed is equal to the molar mass of zinc multiplied by the number of moles of zinc:
0.0051 mol x 65 g/mol = 0.33 grams
So, you would need approximately 0.33 grams of zinc to produce 115 mL of hydrogen gas at STP.