Problem: With 0.14 mol of Mg(s) and 0.020 mol P4(s), what is the number of moles of Mg3P2(s) produced? 6Mg(s) + P4(s) → 2Mg3P2(s)A. 0.004 B. 0.1 C. 0.03 D. 4 x 10-2 E. None of the above

FREE Expert Solution
  • Determine the moles of Mg3Pproduced from 0.14 mole Mg and 0.020 mole P4 via stoichiometry using the balanced equation
  • For this problem, we have to determine which is the limiting reactant among Mg and P4
  • Recall that the limiting reactant will produce the least but correct amount of products
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Problem Details

With 0.14 mol of Mg(s) and 0.020 mol P4(s), what is the number of moles of Mg3P2(s) produced? 

6Mg(s) + P4(s) → 2Mg3P2(s)

A. 0.004 

B. 0.1 

C. 0.03 

D. 4 x 10-2 

E. None of the above

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