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3 years ago

What mass in grams of hydrogen is produced by the reaction of 2.0 g of magnesium? (Make sure to balance the reaction first)

Chemistry

1 answer:

denpristay [2]3 years ago

7 0

Answer:

0.164541341 g H2

Explanation:

1) Convert grams to moles by dividing by RMM of Magnesium (24.31g).

2g Mg * (1 mol Mg / 24.31 g Mg) = 0.082270671 mol of Mg

2) Use the balanced equation's ratio of 1 mol Mg: 1 mol H2.

0.082270671 mol of Mg = 0.082270671 mol of H2

3) Convert the mol of H2 back into grams by multiplying by H2's RMM (2 g).

0.082270671 mol of H2 * 2 g H2 = 0.164541341 g H2

* Answer can be rounded to your liking *

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Step 1: Data given

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Step 2: The balanced equation

2NH3(aq) + CO2(aq) --> CH4N2O(aq) + H2O(l)

Step 3: Calculate moles of NH3

Number of moles = mass / Molar mass

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Moles CO2 = 4803.45 moles

Step 5: Calculate limiting reactant

For 2 moles NH3 consumed, we need 1 moles of CO2 to produce 1 mole urea and 1 mole H2O

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For 7192.77 moles of NH3, we have 3596.4 moles of urea produced

Step 7: Calculate mass of urea

Mass urea = moles urea * molar mass urea

Mass urea = 3596.4 moles * 60.06 g/mol

Mass urea = 216000 grams = 216 kg = theoretical yield

Step 8: Calculate % yield

% yield = (actual yield / theoretical yield)*100%

% yield = (170.3 / 216) *100% = 78.8%

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