when 25.0 g of Zn reacts how many L of H2 gas are formed at STP
Reacts with what? I'll assume HCl but it could be anything.
1. Write the equation.
Zn + 2HCl ==> ZnCl2 + H2
2. Convert 25.0 g Zn to moles. moles = grams/molar mass.
3. Using the coefficients in the balancd equation, convert moles Zn to moles H2.
4. Now convert moles H2 to L.
L = mols H2 x (22.4 L/mol) = ??
To determine the number of liters of H2 gas formed when 25.0 g of Zn reacts at STP, we need to use the stoichiometry of the balanced chemical equation for the reaction.
The balanced equation for the reaction between Zn and H2 gas is:
Zn + 2H2O -> Zn(OH)2 + H2
From the equation, we can see that 1 mole of Zn reacts to produce 1 mole of H2 gas. First, we need to convert the mass of Zn (given as 25.0 g) into moles.
The molar mass of Zn is 65.38 g/mol. We can use this value to convert grams of Zn to moles:
25.0 g Zn × (1 mol Zn / 65.38 g Zn) = 0.382 mol Zn
Now that we have determined the number of moles of Zn, we can directly relate it to the number of moles of H2 gas using the stoichiometry of the balanced equation.
According to the stoichiometry, 1 mole of Zn reacts to produce 1 mole of H2 gas. Therefore, the number of moles of H2 gas formed is also 0.382 mol.
Now, we need to convert moles of H2 gas to volume at STP (Standard Temperature and Pressure). At STP, 1 mole of any gas occupies a volume of 22.4 liters.
0.382 mol H2 × (22.4 L H2 / 1 mol H2) = 8.56 L H2
Therefore, when 25.0 g of Zn reacts, approximately 8.56 liters of H2 gas are formed at STP.