Find out the volume of 6.023*10²² of ammonia at stp?

Question

Henry Antrim · Accepted Answer

The volume is 2.271 L.

Method 1. Using the Ideal Gas Law We can use the Ideal Gas Law to solve this problem. #color(blue)(bar(ul(|color(white)(a/a)pV = nRTcolor(white)(a/a)|)))" "# where We can rearrange the Ideal Gas Law to get #V = (nRT)/p# Step 1. Calculate the moles of ammonia #n = 6.023 × 10^22 color(red)(cancel(color(black)("molecules NH"_3))) × "1 mol NH"_3/(6.022 × 10^23 color(red)(cancel(color(black)("molecules NH"_3)))) = "0.100 02 mol NH"_3# Step 2. Calculate the volume at STP Remember that STP is defined as 0 °C and 1 bar. #n = "0.100 02 mol"# #R = "0.083 14 bar·L·K"^"-1""mol"^"-1"# #T = "(0 + 273.15) K" = "273.15 K"# #p = "1 bar"# #V = (nRT)/p = ("0.100 02" color(red)(cancel(color(black)("g"))) × "0.083 14" color(red)(cancel(color(black)("bar"))) ·"L"· color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 273.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("bar")))) = "2.271 L"# Method 2. Using the molar volume We know that there are 0.100 02 mol of #"NH"_3#. We also know that the molar volume of a gas is 22.71 L at STP. ∴ #V = "0.100 02" color(red)(cancel(color(black)("mol NH"_3))) × "22.71 L NH"_3/(1 color(red)(cancel(color(black)("mol NH"_3)))) = "2.271 L NH"_3#

Noah Aldrich · Answer

undefined Using the ideal gas law, $PV = nRT$, where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the ideal gas constant, and $T$ is temperature. At STP, pressure ($P$) is 1 atm, and temperature ($T$) is 273.15 K.

Rearrange the formula to solve for volume ($V$): $V = \frac{nRT}{P}$. Substitute the given values ($n = 6.023 \times 10^{22}$) and solve.

The volume of $6.023 \times 10^{22}$ moles of ammonia at STP is approximately 22.41 L.