what volume of hydrogen is formed when 3.00g of magnesium react with an excess of dilute sulfuric acid?
Follow these steps.
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To determine the volume of hydrogen formed when 3.00 grams of magnesium reacts with an excess of dilute sulfuric acid, we need to use the molar mass of magnesium and the balanced chemical equation for the reaction.
First, let's find the moles of magnesium using its molar mass. The molar mass of magnesium (Mg) is 24.31 g/mol.
moles of magnesium = mass / molar mass
moles of magnesium = 3.00 g / 24.31 g/mol
Next, we need to use the balanced chemical equation to determine the molar ratio between magnesium and hydrogen. The balanced equation for the reaction between magnesium and sulfuric acid is:
Mg + H2SO4 → MgSO4 + H2
From the balanced equation, we can see that for every 1 mole of magnesium, 1 mole of hydrogen is produced.
Finally, we can calculate the volume of hydrogen using the ideal gas law:
PV = nRT
Assuming the reaction is carried out at standard temperature and pressure (STP), we can use the following values:
- Pressure (P) is 1 atm
- Temperature (T) is 273 K
- Gas constant (R) is 0.0821 L·atm/(mol·K)
n = moles of hydrogen (which is equal to moles of magnesium)
Now, plug the values into the ideal gas law:
V = nRT / P
V = (moles of magnesium) * (0.0821 L·atm/(mol·K)) * (273 K) / (1 atm)
Calculate the moles of magnesium using the formula:
moles of magnesium = 3.00 g / 24.31 g/mol
Finally, substitute the value of moles of magnesium into the ideal gas law equation and calculate the volume of hydrogen.