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 6

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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Reply 7

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

It is really simply 94 not 94.06

2H2 + O2 = 2H2O We’re given the quantity of the reactant for use for this response. These values would be the place to begin of the calculations. First, we have to positive the limiting reactant. 10.54 g H2 ( 1 mol H2 / 2.02 g H2 ) = 5.22 mol H295.10 g O2 ( 1 mol O2 / 32 g H2 ) = 2.97 mol O2 The limiting reactant is the hydrogen fuel. We use this worth for additional calculations. 5.22 mol H2 ( 2 mol H2O / 1 mol H2 ) ( 18.02 g / 1 mol ) = 188.13 g H2O

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It’s important to divide them by one another

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