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Phosphorus combines with hydrogen to form phosphine. In this reaction, 123.9 g of phosphorus combines with excess hydrogen to produce 129.9 g of phosphine. After the reaction, 310 g of hydrogen remains unreacted. What mass of hydrogen is used in the reaction? What was the initial mass of hydrogen before the reaction?​

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There was 321.88 g of hydrogen before the reaction.

The equation of the reaction is;

P4 + 6H2 -------->4PH3

Number of moles of P4 = 123.9 g/124g/mol = 0.99 moles of P4

If 1 mole of P4 reacts with 6 moles of H2

0.99 moles of P4 reacts with x moles of H2

x =  6 moles × 0.99 moles

x = 5.94 moles of H2

Mass of H2 reacted =  5.94 moles of H2 x 2 g/mol

Mass of H2 reacted = 11.88 g of H2

Mass of H2 before the reaction = mass of hydrogen unreacted + mass of hydrogen reacted

Mass of H2 before the reaction = 11.88 g of H2 + 310 g of H2

= 321.88 g

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