1.500 grams of solid magnesium metal were reacted with 70.00 mL of 1.000 M HCl.

Question

1.500 grams of solid magnesium metal were reacted with 70.00 mL of 1.000 M HCl. The resulting H_2 (g) was collected in a gas burette. Mg(s) + 2 HCI (aq) rightarrow MgCl_2(aq) + H_2 (g) a. Determine the limiting reactant in the reaction and show your work. b. Calculate the moles of H_2(g) produced into the gas burette. c. Using the ideal gas law, PV=nRT, calculate the volume (in mL) occupied in the burette by H_2(g). Assume a room temperature of 21.0 degree C and a pressure of 0.8652 atm. The gas constant is R = 0.08206 (L atm)/(mol K).

Explanation / Answer

Balanced equation:
Mg + 2 HCl = MgCl2 + H2

1.5 gm of Mg = 1.5 / 24.305 = 0.06171 Moles

Moles of HCl = 70 x 1 /1000 = 0.07 Moles

One equivalent of Mg needs 2 equivalent of HCl to give product. According to the equation 0.06171 Moles of Mg needs 0.1234 Moles of HCLl . But we have 0.07 Moles of HCl. Hence HCl is the limiting reagent.

Moles of H2 produced = 0.07 /2 = 0.035 Moles

Volume of HCl V = nRT / P

P = 0.8652atm

n = 0.035 Mole

R = 0.0821 L atm K-1 Mole -1

T = 273 + 21 = 294 K

Substitute the values and find volume

V= 0.035 x 0.0821 x 294 / 0.8652 = 0.976 Liter or 976 ml

Volume of H2 = 0.976 Liter or 976 ml

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