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 Mg3P2 produced 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
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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