Equation- Mg + 2HCl -> MgCl2 + H2

What volume of hydrogen at STP is produced from the reaction of 50.0 g of Mg and the equivalent of 75g of HCl?

To get the volume of hydrogen, you need to know how many MOLES of hydrogen are produced. To know this, you need to know how many moles of reactants there are. Do you know how to do that?

yes

OK. So, calculate how many moles of each reactant is present...and then, by youre balanced equation, you know that one mole of Mg forms one mole of H2...it's 1 to 1.

Woops. Yikes! I left something out. Don't forget that twice as many moles of HCl than Mg is required....so, when you do the calculation you see that there are 2.05 moles of HCl and 2.05 moles of Mg present. The Mg all gets used up with half left over...so it's 1.025 moles of Mg + 2.05 moles of HCL goes to 1.025 moles of MGCl2 + 1.025 moles of H2. Sorry if I mislead you by leaving this out!!

.13

No problem! Let's continue with the correct information.

To calculate the volume of hydrogen gas produced at STP (Standard Temperature and Pressure), we can use the equation:

1 mole of any gas at STP = 22.4 liters.

From the balanced equation:
1 mole of Mg produces 1 mole of H2.

So, 1.025 moles of Mg will produce 1.025 moles of H2.

Therefore, the volume of hydrogen gas produced at STP can be calculated as:

1.025 moles H2 x 22.4 L/mole = 22.92 liters of H2.

So, 22.92 liters of hydrogen gas will be produced from the reaction of 50.0g of Mg and the equivalent of 75g of HCl at STP.

No problem! Thank you for catching that. Now that we have the correct mole ratios, we can calculate the volume of hydrogen gas produced at STP.

First, we need to convert the moles of hydrogen gas to volume using the ideal gas law. The ideal gas law equation is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

At STP (Standard Temperature and Pressure), the pressure is 1 atm and the temperature is 273.15 K.

We know that 1 mole of any gas occupies 22.4 liters at STP. So, we can use this conversion factor to calculate the volume of hydrogen gas.

The balanced equation tells us that 1.025 moles of magnesium (Mg) reacts to produce 1.025 moles of hydrogen gas (H2).

Therefore, the volume of hydrogen gas produced can be calculated as follows:

V(H2) = n(H2) x 22.4 L/mol

V(H2) = 1.025 mol x 22.4 L/mol

V(H2) = 22.92 L

So, the volume of hydrogen gas produced at STP from the reaction of 50.0 g of Mg and the equivalent of 75 g of HCl is 22.92 L.