When aqueous solutions of Li3PO4 and MgSO4 are mixed, Mg3(PO4)2 is a precipitate
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Answer:
0.4 g of hydrogen gas would be produced.
1. 0.48 mole of HCl is needed to react completely with 15.5 g of zinc.
2. HCl is the limiting reactant.
3. 2.61 g of zinc is in excess
Explanation:
From the balanced equation of reaction:
1 mole of Zn requires 2 moles of HCl to produce 1 mole of hydrogen gas.
15.5 g of zinc = 15.5/65.3 = 0.24 moles of zinc.
0.24 moles Zn is supposed to require 0.24 x 2 moles HCl which is equivalent to 0.48 moles HCl.
But only 0.400 mole of HCl is present. <u>Hence, HCl is the limiting reagent</u>. <u>For complete reaction with 15.5 g of Zinc, 0.48 mole HCl would be needed.</u>
0.400 mole of HCl will require 0.2 mole of Zn for complete reaction. <u>This thus means that 0.24 - 02 = 0.04 mole of Zn is in excess.</u>
0.04 mole Zn = 0.04 x 65.3 = 2.61 g excess Zn.
Now, since HCl is the limiting reagent;
2 moles of HCl is required to produce 1 mole of H2 according to the equation.
0.400 mole HCl will therefore yield 0.400 x 1/2 = 0.2 mole H2
0.2 mole H2 = 2 x 0.2 = 0.4 g H2
<em>Hence, </em><em>0.4 g</em><em> of hydrogen gas would be produced.</em>