_H2SO4 + __ B(OH)3 = _ B2(SO4)3 + __ H2O
1 answer:
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Answer:
12.5 g
Explanation:
<em>12.5 g of the compound would be formed.</em>
First, let us look at the balanced equation of reaction.
3 moles of Mg is required to react with 1 mole of N2 to produce 1 mole of product.
<em>Recall that: mole = mass/molar mass</em>
9.03 g of Mg = 9.03/24.3 = 0.3716 mole
3.48 g of N2 = 3.48/28 = 0.1243 mole
Mole ratio of Mg/N2 = 3:1
<em>Hence, there is no limiting reactant.</em>
3 moles of Mg is required for 1 mole of product.
0.3716 mole of Mg will therefore require:
0.3716 x 1/3 = 0.1239 moles of product.
Molar mass of product = 100.9 g/mol
Mass of 0.1239 mole = mole x molar mass
= 0.1239 x 100.9 = 12.5 g