Suppose 8.00 G Of Ch4
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Suppose viii.50 g of CH4 is allowed to burn in the presence of 16.70 thousand of oxygen. How much (in grams) CH4, O2, CO2, and H2O remain afterwards the reaction is consummate?
Related Question
Consider the reaction CH4(1000) + 2O2(grand)—> CO2(g)+2H2O(1000) how many grams of methyl hydride should be burned in an excess of oxygen at STP to obtain five.half-dozen L of carbon dioxide? 2.0 thousand iv.0 one thousand xvi.0 g 32.0 thou
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Video Transcript
we are given this conversion reaction of the methyl hydride and we are given that the volume produced of the C. 02 gas, allow that exist Five. That is given as a 5.6 liter at STP and we need to find the mask of the methane required to produce v.6 L of the carbon dioxide gas. And so showtime of all, we will find the number of moles of c. 0. 2, that volition be equal to the volume of C. 02, which is 5.six divide 22.four. That comes out to be around 0.25 mol of the C. 02. Because we know that at STP i mole of any gas occupies 22.4 liter volume. At present nosotros will find the number of moles of methane number of moles of C H 4. That will be equal to wait at this reaction here, one mol ch four produces one mole of C. 02. That means the number of moles are same. And so here we can write number of moles of CS iv is equal to 0.25. Now, we can easily find the mask of CH 4, that will be equal to the number of moles. Multiply the tooth mass number of moles is 0.25. Multiply the tooth mass of the methane is sixteen thousand per mole. And so we get around iv 1000 method is required
Suppose 8.00 G Of Ch4,
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