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What is the volume of 19.87mol of ammonium chloride NH4CL at STP
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What is the volume of 19.87mol of ammonium chloride NH4CL at STP

User Sander Roest
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n = (V)/(STP) \\ n = (V)/(22.4) \\ 19.87 = (V)/(22.4) \\ V = 19.87 * 22.4 \\ V = 445,088 \: l \: of \: NH4CL \:

User Fastkowy
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