Question

when 1.10 g of magnesium reacts with 300.0mL of 0.800 M HCL,the products are hydrogen gas and magnesium chloride . 2HCL + Mg a MgCl2 +H2
a) Determine the moles of hydrogen gas produced(assume Mg is limiting reagent)
b) Determine the moles of hydrogen gas at STP
C) The reaction, was actually carried out at room temperature,25.0 C.What was the volume of hydrogen gas produced under these conditions?

Answer 1

Answer on Question 83018 in General Chemistry

.m (Mg) = 1.10 g

V (HCl sol) = 300 mL

.c (HCl) = 0.800 M

2HCl + Mg = MgCl₂ + H₂

.a) n (H₂) = ?

.b) n (H₂) at STP = ?

.c) V (H₂) at 25 °C = ?

.a) Find the amount of substance of Mg n = $\frac{m}{Ar}$ = $\frac{1.1}{24}$ = 0.046 mol

Find the amount of substance of HCl n = C × V = 0.800 × 0.3 = 0.024 mol

Mg is limiting reagent n (H₂) = n (Mg) = 0.046 mol

.b) find the volume of H₂ at STP V = V_M × n = 22.4 × 0.046 = 1.03 L

The moles of hydrogen gas at STP have the same value that is received in a)

.c) to find the volume at 25°C or 298 K we use Gay Lussac law (p = const)

$\frac{V_2}{T_1} = \frac{V_2}{T_2}$ from which $V_2 = \frac{V_1 \times T_2}{T_1} = \frac{1.03 \times 298}{273} = 1.12$ L

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