# How many grams of water will form if 10.54 g H2 react with 95.10 g O2?

What number of grams of water will kind if 10.54 g H2 react with 95.10 g O2?

94.0 g H2O Clarification:

94.0 g H2O Clarification: 2H2 + O2 –> 2H2O The query requested me what number of grams of water will kind if 10.54 g H2 react with 95.10 g O2. The right reply was 94.0g of H2O I acquired from 10.54g H2 * 1 mol H2/2.02 g/mol H2 * 2 mol H2O/2 mol H2 = 5.22 mol H2O 95.10g O2* 1 mol O2/32 g/mol O2 * 2 mol H2O/1 mol O2 = 5.94 mol H2O The limiting reactant is 5.22 mol, which 5.22 mol H2O * 18.01 g/mol H2O = 94.0g H2O.

m(H₂)=10.54 gM(H₂)=2.016 g/molm(O₂)=95.10 gM(O₂)=31.999 g/molM(H₂O)=18.015 g/mol n(H₂)=m(H₂)/M(H₂)n(H₂)=10.54/2.016=5.228 moln(O₂)=m(O₂)/M(O₂)n(O₂)=95.10/31.999=2.972 mol H₂:O₂=2:15.288:2.972 ⇒ 2:1.124    oxygen in extra n(H₂O)=n(H₂)m(H₂O)=n(H₂)M(H₂O)m(H₂O)=5.228*18.015=94.182 g

94.0 g H2O is the right reply I simply did it

94.0 Clarification: 2020 edge

Reply : The quantity of water might be, 94.86 grams Resolution : First we’ve got to calculate the moles of and . Now we’ve got to calculate the limiting and extra reagent. The balanced chemical response is, From the balanced response we conclude that As, 2 moles of react with 1 mole of So, 5.27 moles of react with moles of Which means, within the given balanced response, is a limiting reagent as a result of it limits the formation of merchandise and is an extra reagent. The surplus reagent stays  = 2.97 – 2.63 = 0.34 moles Now we’ve got to calculate the moles of . As, 2 moles of react with 2 moles of So, 5.27 moles of react with moles of Now we’ve got to calculate the mass of . Due to this fact, the quantity of water might be, 94.86 grams

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