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Sagot :
Answer:
25.8g of NH3 can be formed
Explanation:
To solve this question we must convert the mass of each reactant to moles using molar mass. With the chemical equation we can find limiting reactant. The moles of limiting reactant are used yo find the moles of ammonia produced and its mass:
Moles NO -30.01g/mol-
45.4g * (1mol / 30.01g) = 1.51 moles
Moles H2 -Molar mass: 2.01g/mol-
12.1g * (1mol / 2.01g) = 6.02 moles
For a complete reaction of 6.02 moles of H2 are needed:
6.02mol H2 * (2mol NO / 5mol H2) = 2.41 moles NO are needed
As there are just 1.51 moles, NO is limiting reactant
Moles NH3:
1.51 moles NO * (2mol NH3 / 2mol NO) = 1.51 moles NH3
Mass NH3 -Molar mass: 17.031g/mol-
1.51 moles NH3 * (17.031g / mol) =
25.8g of NH3 can be formed
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