How many moles of H3PO4 are produced when 71 g of P4010 reacts completely? P4010(s) + 6H,0()- 4H3PO4(aq) O a 0.0635 moles H3PO4 O b 16.0 moles H3PO4 4.00 moles H3PO4 O d 1.00 moles H3PO4

General, Organic, and Biological Chemistry
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Chapter6: Chemical Calculations: Formula Masses, Moles, And Chemical Equations
Section: Chapter Questions
Problem 6.71EP: How many moles of beryllium (Be) are needed to completely react with 10.0 moles of N2 in the...
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How many moles of H3PO4 are produced when 71 g of P4010 reacts completely
P4010(s) + 6H20() → 4H3PO4(aq)
O a
0.0635 moles H3PO4
O b
16.0 moles H3PO4
O c
4.00 moles H3PO4
O d
1.00 moles H3PO4
Transcribed Image Text:How many moles of H3PO4 are produced when 71 g of P4010 reacts completely P4010(s) + 6H20() → 4H3PO4(aq) O a 0.0635 moles H3PO4 O b 16.0 moles H3PO4 O c 4.00 moles H3PO4 O d 1.00 moles H3PO4
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