Question

4.What volume of hydrogen gas at STP is produced when 2.5 grams of zinc react with an

excess of hydrochloric acid?

Answer 1

0.86

Step-by-step explanation:

1mol of Zn has mass of 65.39g.The amount of Zn is 2.5g65.39g/mol=0.038mol.

The amount of H2 produced is the same as the amount of Zn consumed (0.038mol).

1mol of ideal gas will occupy 22.4L at STP.

The H2 will occupy 0.038mol×22.4L/mol=0.86L

.

Answer 2

When 2.5 grams of zinc reacts with excess hydrochloric acid, the resulting hydrogen gas at STP is approximately 0.0328 liters.

To determine the volume of hydrogen gas produced when 2.5 grams of zinc reacts with excess hydrochloric acid, we utilize the stoichiometry of the chemical equation and the molar volume of a gas at standard temperature and pressure (STP), which is 0.86 L/mol.

First, we find the moles of zinc by dividing the given mass (2.5 grams) by the molar mass of zinc (approximately 65.38 g/mol). The result is approximately 0.038 moles of zinc.

Next, using the balanced chemical equation
`\( \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \)`, we see that one mole of zinc produces one mole of hydrogen gas. Therefore, 0.038 moles of zinc will produce 0.038 moles of hydrogen gas.

Finally, applying the ideal gas law at STP
`(\(0.86 \, \text{L/mol}\))`, we calculate the volume of hydrogen gas:

`\[ \text{Volume of H}_2 = \text{Number of moles of H}_2 * \text{Molar volume at STP} \]`

`\[ \text{Volume of H}_2 = 0.038 \, \text{moles} * 0.86 \, \text{L/mol} \]`

`\[ \text{Volume of H}_2 \approx 0.0328 \, \text{L} \]`

Therefore, 2.5 grams of zinc reacting with excess hydrochloric acid produces approximately 0.0328 liters of hydrogen gas at STP.