Question

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

Answer 1

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

$m=M\times n=64.044\times0.0383=2.45 g$