7.5 L of hydrogen gas at STP and zinc (II) nitrate was produced by reacting zinc with nitric acid. Calculate the mass of zinc needed for this reaction.
This is what I did: Zn + 2HNO3 = H2 + Zn(NO3)2.
7.5L H2 * 1 mole H2/ 22.4 L H2 * 1 mole Zn/ 1 mole H2 * 65.4 g Zn / 1 mole Zn
I got 21.9 g Zn but I am not sure it is right.
That looks good to me but you have too many significant figures if you posted the problem correctly. 7.5 L allows only two places and I would round to 22 g.
Well, your calculations seem to be on the right track. You correctly used the molar ratios from the balanced equation. However, when calculating the mass of zinc needed, you need to use the molar mass of zinc (65.4 g/mol), not the molar mass of hydrogen (1 g/mol).
So, let's recalculate it properly:
7.5 L H2 * 1 mole H2 / 22.4 L H2 * 1 mole Zn / 1 mole H2 * 65.4 g Zn / 1 mole Zn
By canceling the units correctly, we get:
(7.5 L/22.4) * (1/1) * (65.4 g/1)
This gives us approximately 17.37 g Zn.
Therefore, the mass of zinc needed for this reaction is around 17.37 grams.
Your calculation is correct! Well done!
To calculate the mass of zinc needed for the reaction, you can use the balanced chemical equation:
Zn + 2HNO3 -> H2 + Zn(NO3)2
First, convert the given volume of hydrogen gas (7.5 L) at STP to moles using the ideal gas law:
7.5 L H2 * (1 mol H2/22.4 L H2) = 0.3354 mol H2
Next, use the stoichiometry of the balanced equation to relate the moles of hydrogen gas to moles of zinc:
0.3354 mol H2 * (1 mol Zn/1 mol H2) = 0.3354 mol Zn
Finally, convert the moles of zinc to grams using the molar mass of zinc:
0.3354 mol Zn * (65.4 g Zn/1 mol Zn) = 21.9 g Zn
Therefore, the mass of zinc needed for this reaction is 21.9 grams.
To calculate the mass of zinc needed for the reaction, you can use the stoichiometry of the balanced equation as you did. Here is a step-by-step explanation of the calculation:
1. Start with the given volume of hydrogen gas at STP: 7.5 L.
Note: STP stands for standard temperature and pressure, which is 0 degrees Celsius (273.15 K) and 1 atmosphere (atm) pressure.
2. Use the molar volume of an ideal gas at STP, which is 22.4 L/mol. This gives you the number of moles of hydrogen gas:
7.5 L H2 * (1 mol H2 / 22.4 L H2) = 0.335 mol H2
3. Next, use the stoichiometric ratio from the balanced equation to convert moles of hydrogen gas to moles of zinc:
0.335 mol H2 * (1 mol Zn / 1 mol H2) = 0.335 mol Zn
4. Finally, convert moles of zinc to grams using the molar mass of zinc (65.4 g/mol):
0.335 mol Zn * (65.4 g Zn / 1 mol Zn) = 21.9 g Zn
So, your calculation of 21.9 g Zn is correct. The mass of zinc needed to produce 7.5 L of hydrogen gas at STP in this reaction would be 21.9 grams.