# How many grams of water will form if 10.54 g H2 react with 95.10 g O2 [Solved] 2H2 O2 = 2H2O Were 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

It is really simply 94 not 94.06

94.0 Clarification: 2020 edge

Reply : The quantity of water might be, 94.86 grams Resolution : First weve got to calculate the moles of and . Now weve 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 weve 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 weve got to calculate the mass of . Due to this fact, the quantity of water might be, 94.86 grams

Reply : The quantity of water might be, 94.86 grams Resolution : First weve got to calculate the moles of and . Now weve 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 weve 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 weve got to calculate the mass of . Due to this fact, the quantity of water might be, 94.86 grams

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Reply : The quantity of water might be, 94.86 grams Resolution : First weve got to calculate the moles of and . Now weve 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 weve 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 weve got to calculate the mass of . Due to this fact, the quantity of water might be, 94.86 grams

2H2 O2 = 2H2O Were 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

It is really simply 94 not 94.06

Reply : The quantity of water might be, 94.86 grams Resolution : First weve got to calculate the moles of and . Now weve 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 weve 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 weve got to calculate the mass of . Due to this fact, the quantity of water might be, 94.86 grams

Get Answer for  Which of the following compounds has the same percent composition by mass as styrene, C8H8 [Solved]

Reply : The quantity of water might be, 94.86 grams Resolution : First weve got to calculate the moles of and . Now weve 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 weve 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 weve got to calculate the mass of . Due to this fact, the quantity of water might be, 94.86 grams

2H2 O2 = 2H2O Were 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

94.0 Clarification: 2020 edge

Its important to divide them by one another

This post is last updated on hrtanswers.com at Date : 1st of September – 2022