Ydrogen and fluorine combine according to the equation h2(g) + f2(g) → 2 hf(g) if 5.00 g of hydrogen gas are combined with 38.0 g of fluorine gas, the maximum mass of hydrogen fluoride that could be produced is
1 answer:
<span> Molar mass (H2)=2*1.0=2.0 g/mol Molar mass (F2)=2*19.0=38.0 g/mol Molar mass (HF)=1.0+19.0=20.0 g/mol 5.00 g H2 * 1mol H2 /2 g H2=2.50 mol H2 38.0 g F2*1mol F2/38.0 g F2=1.00 mol F2 H2(g) + F2(g) → 2 HF(g) From reaction 1 mol 1 mol From problem 2.50 mol 1 .00mol We can see that excess of H2, and that F2 is a limiting reactant. So, the amount of HF is limited by the amount of F2. </span> H2(g) + F2(g) → 2 HF(g) From reaction 1 mol 2 mol From problem 1.00 mol 2.00mol 2.00 mol HF can be formed. 2.00 mol HF*20.0g HF/1mol HF=40.0 g HF can be formed
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