of will be produced from 25.2 g of and 8.42 g of .
Further explanation:
Limiting reagent:
It is completely consumed in a chemical reaction. It decides the amount of product formed in any chemical reaction. The amount of product depends on the amount of limiting reagent since the product formation is not possible in the absence of it.
The given balanced chemical equation for the formation of is as follows:
From the balanced chemical reaction, the reaction stoichiometry between and is as follows:
According to the reaction stoichiometry, 3 moles of and 1 mole of react to form 2 moles of but the reaction mixture has 25.2 g of and 8.42 g of .
The formula to calculate the moles of is as follows:
...... (1)
Substitute 8.42 g for the given mass of and 2.02 g/mol for the molar mass of in equation (1).
The formula to calculate the moles of is as follows:
...... (2)
Substitute 25.2 g for the given mass of and 28.01 g/mol for the molar mass of in equation (2).
Calculate the moles of that react with and 4.168 moles of as follows:
The calculations concluded that 1.389 moles of is required for 4.168 moles of but we have only 0.899 moles of in the reaction mixture. Therefore, is present in limited quantity and is a limiting reagent.
According to the balanced chemical equation 1 mole of produce 2 moles of , therefore, the moles of produced by 1.389 moles of can be calculated as follows:
The formula to calculate the mass of is as follows:
...... (3)
Substitute 1.798 mol for the moles of and 17.03 g/mol for the molar mass of in equation (3).
So 30.62 g of will be produced from 25.2 g of and 8.42 g of .
Learn more:
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Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Mole concept
Keywords: limiting reagent,NH3,N2,H2,2NH3,3H2,30.62 g,2.778 mol,17.03 g/mol,2.02 g/mol,1.389 mol,25.2 g,8.42 g,4.168 mol,0.899 mol,1 mole,2 moles,3 moles.