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Answer:
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The mass of hydrogen in 57.010 g ammonium hydrogen phosphate, (NH₄)H₂PO₄ is 2.97 g
<h3>Determination of mass of 1 mole of (NH₄)H₂PO₄ </h3>
1 mole of (NH₄)H₂PO₄ = 14 + (4×1) + (2×1) + 31 + (16×4) = 115 g
<h3>Determination of the mass of H in 1 mole of (NH₄)H₂PO₄ </h3>
Mass of H = 6H = 6 × 1 = 6 g
<h3>Determination of the mass of H in 57.010 g of (NH₄)H₂PO₄ </h3>
115 g of (NH₄)H₂PO₄ contains 6 g of H.
Therefore,
57.010 g of (NH₄)H₂PO₄ will contain = (57.010 × 6) / 115 = 2.97 g of H
Thus, 2.97 g of Hydrogen, H is present in 57.010 g of (NH₄)H₂PO₄
Learn more about mass composition:
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