Calculate the mass of H_3PO_4 that could possibly be produced by the reaction of

Question

Calculate the mass of H_3PO_4 that could possibly be produced by the reaction of 300 mg of P_4O_10 with 100 mg of water. P_4O_10 + 6 H_2O rightarrow 4 H_3PO_4 Identify the limiting reagent. A chemist isolated 25.2 g of B_2O_3, from the reaction of 25.0g of boron with an excess of oxygen. Calculate the percent yield of B_2O_3, 4B + 3O_2 rightarrow 2B_2O_3 19.2 g of tungsten is isolate from the reaction of 25.00 g of WO_3 and 2.00 g of H_2. What is the percent yield of tungsten? WO_3 + 3H_2 rightarrow 3H_2O

Explanation / Answer

24 )

P4O10   + 6H2O ----------------> 4 H3PO4

283.9g         108 g                          392 g

0.30 g           0.1 g                          ?

283.9 g P4O10 needes ------------------> 108 g H2O

0.30 g P4O10 needed --------------------> 108 x 0.30 / 283.9 = 0.114 g H2O needed

but but we have only 0.1 g H2O . so

H2O is limiting reagent

mass of H3PO4 = 0.1 x 392 / 108

  mass of H3PO4     = 0.363 g

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