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Zorluk: OrtaChemical Equations and Stoichiometric Calculations (Mass-Mass, Mass-Volume, Mole Relations)

What volume of hydrogen gas, measured at STP, is produced when 4.8 g4.8\text{ g} of magnesium ribbon reacts completely with excess dilute tetraoxosulfate(VI) acid according to the chemical equation below?

Mg(s)+H2SO4(aq)→MgSO4(aq)+H2(g)Mg(s) + H_2SO_4(aq) \rightarrow MgSO_4(aq) + H_2(g)

[Mg=24,Molar volume of gas at STP=22.4 dm3mol−1][Mg = 24, \text{Molar volume of gas at STP} = 22.4\text{ dm}^3\text{mol}^{-1}]

Cevap: 4.48 dm^3

Cevap

The volume of hydrogen gas produced at STP is 4.48 dm34.48\text{ dm}^3.
From the stoichiometric relationship in the balanced reaction Mg(s)+H2SO4(aq)→MgSO4(aq)+H2(g)Mg(s) + H_2SO_4(aq) \rightarrow MgSO_4(aq) + H_2(g), 1 mol1\text{ mol} of MgMg (24 g24\text{ g}) produces 1 mol1\text{ mol} of H2H_2 gas (22.4 dm322.4\text{ dm}^3 at STP). For 4.8 g4.8\text{ g} of MgMg, the number of moles is 4.824=0.20 mol\frac{4.8}{24} = 0.20\text{ mol}. Multiplying by the molar gas volume gives 0.20×22.4=4.48 dm30.20 \times 22.4 = 4.48\text{ dm}^3 of H2H_2 gas.

Adım Adım Çözüm

1
Calculate the amount in moles of magnesium (MgMg) reacted.
n(Mg)=4.8 g24 g mol−1=0.20 moln(Mg) = \frac{4.8\text{ g}}{24\text{ g mol}^{-1}} = 0.20\text{ mol}
Dividing given mass by relative atomic mass yields the number of moles.
2
Determine the amount in moles of hydrogen gas (H2H_2) produced.
n(H2)=0.20 moln(H_2) = 0.20\text{ mol}
The balanced chemical equation shows a 1:11:1 stoichiometric molar ratio between MgMg and H2H_2.
3
Calculate the volume of H2H_2 gas at STP.
V(H2)=0.20 mol×22.4 dm3mol−1=4.48 dm3V(H_2) = 0.20\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 4.48\text{ dm}^3
At standard temperature and pressure (STP), one mole of any gas occupies 22.4 dm322.4\text{ dm}^3.

Anahtar Kavram

Mass-Volume stoichiometric calculation at STP
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Chemical Equations and Stoichiometric Calculations (Mass-Mass, Mass-Volume, Mole Relations)

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