Astudent performed an experiment in the same manner as part b (magnesium+hcl). the temperature for the day was 22°c in the barometric pressure was 742mmhg. 43.5 ml of gas was collected how many grams of magnesium must've been used in the reaction?

0.0425 g

We are given;

but, K = °C + 273.15

Thus, T = 295.15 K

We are required to calculate the mass of magnesium that was used in the reaction.

Using the ideal gas equation, we can determine the number of moles

PV = nRT , where R is the ideal gas constant, 62.3637 L·mmHg/mol·K

Rearranging the formula;

n = PV ÷ RT

  = (742 mmHg× 0.0435 L) ÷ (62.3637 × 295.15K)

  = 0.00175 moles

The equation fro the reaction is;

Mg + 2HCl → MgCl₂ + H₂

Thus, 1 mole of Mg reacts to produce 1 mole of hydrogen gas

Therefore, moles of Mg = Moles of Hydrogen gas

                                        = 0.00175 moles

We know that, Moles = Mass ÷ Molar mass

Thus, Mass = Moles × Molar mass

Molar mass of Mg = 24.305 g/mol

Hence;

Mass of Mg = 0.00175 moles × 24.305 g/mol

                   = 0.0425 g

Thus, 0.0425 g of magnesium reacted

if you're asking what comes out, it's steam.