Answer:
The excess reactant is .
Left over = 1.9903 moles or 4.0122 g
Explanation:
The formula for the calculation of moles is shown below:
Given: For
Given mass = 37.0 g
Molar mass of = 28.0134 g/mol
<u>Moles of = 37.0 g / 28.0134 g/mol = 1.3208 moles
</u>
Given: For
Given mass = 12.0 g
Molar mass of = 2.0159 g/mol
<u>Moles of = 12.0 g / 2.0159 g/mol = 5.9527 moles
</u>
According to the given reaction:
1 mole of react with 3 moles of
1.3208 moles of react with 3*1.3208 moles of
Moles of reacted = 3.9624 moles
Available moles of = 5.9527 moles
<u>
Limiting reagent is the one which is present in small amount. Thus, is limiting reagent as is present in excess . (3.9624 < 5.9527)
</u>
<u>The excess reactant is .</u>
Left over = 5.9527 - 3.9624 moles = 1.9903 moles
Mass = moles × Molar mass = 1.9903 moles × 2.0159 g/mol = 4.0122 g
<u>Left over is 4.0122 g of hydrogen.</u>