please with steps
NH4NO2 (s) = N2 (g) + H2O (g)
When a sample is decomposed in a test tube, 949 mL of wet N2(g) is collected over water at 26°C and 778 torr total pressure. How many grams of dry NH4NO2(s) were initially decomposed? The vapor pressure of water at 26°C is 25.2 torr.
Mass=.............g
1.find the amount of substance
"PV = nRT"
P=778-25,2=752.8=0.990526712 atm
26°C=299.15 K
"0.990526712\\times0.949 = n\\times0.08206\\times299.15"
"0.94 = n\\times24.55"
n=0.0383
"m=M\\times n=64.044\\times0.0383=2.45 g"
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