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34. When ammonia gas, oxygen gas and methane gas (ch4) are combined, the products are hydrogen cyanide gas andrnwater. Calculate the mass of each product produced when 2.25x102rnrng of oxygen gas is reacted with an excess ofrnthe other two reactants. if the actual yield of the experiment is 105 g of HCN, calculate the percent yield.
Accepted Answer
The balanced chemical equation for the reaction is:
2 NH3(g) + 3 O2(g) + 2 CH4(g) → 2 HCN(g) + 6 H2O(g)
First, we need to convert the mass of oxygen to moles using its molar mass (32.00 g/mol):
2.25 x 10^2 g O2 x (1 mol O2 / 32.00 g O2) = 7.03 mol O2
Now we can use the stoichiometry of the balanced equation to calculate the moles of each product:
- Moles of HCN: 7.03 mol O2 x (2 mol HCN / 3 mol O2) = 4.69 mol HCN
- Moles of H2O: 7.03 mol O2 x (6 mol H2O / 3 mol O2) = 14.1 mol H2O
Next, convert the moles of each product to grams using their respective molar masses:
- Mass of HCN: 4.69 mol HCN x (27.03 g HCN / 1 mol HCN) = 126 g HCN
- Mass of H2O: 14.1 mol H2O x (18.02 g H2O / 1 mol H2O) = 254 g H2O
Therefore, the theoretical yield of HCN is 126 g.
Finally, calculate the percent yield:
Percent yield = (actual yield / theoretical yield) x 100%
Percent yield = (105 g / 126 g) x 100% = 83.3%
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