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how many grams of PH3 will be produced if 6.0 l of H2 are used?

Sagot :

Mass of PH3= 6.086 g

Further explanation

Given

6.0 L of H2

Required

mass of PH3

Solution

Reaction

P4 + 6H2 → 4PH3

Assumed at STP ( 1 mol gas=22.4 L)

Mol of H2 for 6 L :

= 6 : 22.4 L

= 0.268

From the equation, mol PH3 :

= 4/6 x moles H2

= 4/6 x 0.268

= 0.179

Mass PH3 :

= 0.179 x 33,99758 g/mol

= 6.086 g