Magnesium reacts with dilute hydrochloric acid. The gas given off is collected in a gas syringe, as shown in Fig. 4.1. Fig. 4.1 The equation for the reactio...
Question 1 Report
Magnesium reacts with dilute hydrochloric acid. The gas given off is collected in a gas syringe, as shown in Fig. 4.1.
Fig. 4.1
The equation for the reaction is: Mg + 2HCl → MgCl2 + H2
(a) Give the names of the two products of this reaction. [2]
(b) Calculate the amount, in mol, in 6.0 g of magnesium. Use Ar: Mg = 24. [2]
(c) Use the equation to determine the amount, in mol, of hydrogen produced, then calculate the number of hydrogen molecules formed. Use the Avogadro constant 6.02 × 1023. [4]
(d) Calculate the mass of magnesium chloride formed. Use Ar: Mg = 24, Cl = 35.5. [2]
Answer Details
This question uses a balanced equation to carry out reacting mass and particle number calculations from moles.
(a) Names of the two products [2]. From \(Mg + 2HCl \rightarrow MgCl_2 + H_2\), the products are magnesium chloride [1] and hydrogen [1].
(b) Amount of magnesium in 6.0 g [2]. Using moles = mass ÷ Ar:
\[ \text{amount} = \frac{6.0}{24} = 0.25 \text{ mol} \]the working [1] and the answer 0.25 mol [1].
(c) Moles and number of hydrogen molecules [4]. The equation shows a 1:1 ratio of Mg to H2, so 0.25 mol of Mg gives 0.25 mol of H2 [1] [1]. Then multiply by the Avogadro constant:
\[ 0.25 \times 6.02 \times 10^{23} = 1.505 \times 10^{23} \text{ molecules} \]the substitution [1] and the answer \(1.505 \times 10^{23}\) molecules [1].
(d) Mass of magnesium chloride formed [2]. First Mr(MgCl2) = \(24 + (2 \times 35.5) = 24 + 71 = 95\). The ratio Mg:MgCl2 is 1:1, so 0.25 mol of MgCl2 forms:
\[ \text{mass} = 0.25 \times 95 = 23.75 \text{ g} \]correct Mr and set up [1], answer 23.75 g [1].
Exam tip: always read the mole ratio straight from the balancing numbers in the equation. Here every ratio is 1:1, so the moles of Mg carry through unchanged to H2 and MgCl2.
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