If equal masses of Sodium and Strontium were reacted with excess hydrochloric acid, which experiment would produce a greater volume of hydrogen gas?
To determine which experiment would produce a greater volume of hydrogen gas when equal masses of Sodium and Strontium are reacted with excess hydrochloric acid, we need to compare the stoichiometry and molar ratios in the respective chemical reactions.
Let's start with the chemical equations for the reaction of Sodium and Strontium with hydrochloric acid:
1. Sodium (Na) reacts with hydrochloric acid (HCl):
2Na + 2HCl -> 2NaCl + H2
2. Strontium (Sr) reacts with hydrochloric acid (HCl):
Sr + 2HCl -> SrCl2 + H2
From the balanced equations, we can see that both reactions produce hydrogen gas (H2), but the coefficients in front of the reactants differ.
In the case of Sodium, two moles of Sodium (2Na) reacts with two moles of hydrochloric acid (2HCl) to produce one mole of hydrogen gas (H2). This means that for every 2 grams of Sodium, we get one gram of hydrogen gas.
For Strontium, one mole of Strontium (Sr) reacts with two moles of hydrochloric acid (2HCl) to produce one mole of hydrogen gas (H2). This implies that for every 1 gram of Strontium, we also get one gram of hydrogen gas.
Since the masses of Sodium and Strontium are equal in the given scenario, both experiments will produce an equal volume of hydrogen gas. The volume of hydrogen gas produced is directly proportional to the number of moles, which in this case is identical for both Sodium and Strontium.
Thus, both Sodium and Strontium will yield the same volume of hydrogen gas when reacted with excess hydrochloric acid, assuming the masses of both metals are equal.
To determine which experiment would produce a greater volume of hydrogen gas when equal masses of Sodium and Strontium are reacted with excess hydrochloric acid, we first need to understand the stoichiometry of the reaction.
The balanced chemical equation for the reaction between a metal and hydrochloric acid is:
2M(s) + 2HCl(aq) → 2MCl(aq) + H2(g)
From the equation, we can see that 1 mole of metal (M) reacts with 1 mole of hydrochloric acid (HCl) to produce 1 mole of hydrogen gas (H2).
Since we have equal masses of Sodium and Strontium, we need to compare the molar masses of the two elements to determine the number of moles of each.
The molar mass of Sodium (Na) is approximately 23 g/mol, while the molar mass of Strontium (Sr) is approximately 88 g/mol.
Assuming we have the same mass of both Sodium and Strontium (let's say 1 gram for simplicity), we can calculate the number of moles for each element:
Number of moles of Sodium (Na):
1 g / 23 g/mol ≈ 0.043 moles
Number of moles of Strontium (Sr):
1 g / 88 g/mol ≈ 0.011 moles
Since the stoichiometry of the reaction is 1 mole metal to 1 mole hydrogen gas, we can conclude that the experiment with Sodium would produce a greater volume of hydrogen gas.
To quantify the exact volume of hydrogen gas produced, you would need to use the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. By plugging in the values for moles of Sodium and Strontium, as well as other known parameters, you can calculate the respective volumes of hydrogen gas produced in each experiment.
You have to start with balanced equation..
2Na+2HCl> 2NaCl + H2 or one mole Na gives 1/2 mole of H2
Sr +2HCl>SrCl2 + H2 so one mole of Sr gives one mole of H2
Now mass
let m be the mass
m/23 = moles of Na
m/38=mole Sr
amount of gas:
Sr: m/38
Na: m/23 (1/2)=m/46
well, m/38 is greater than m/46, so the answer is...