Cho 37,10 gam muối cacbonat của kim loại kiềm m tác dụng hết với dung dịch HCl (dư) , sinh ra 7,84 lít khí (đktc). Kim loại M là A. Rb B. Li C. Na D. K
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Using the equation M2CO3 + 2HCl -> 2MCl + CO2 + H2O, we can determine that the mole ratio between the carbonate salt and HCl is 1:2. Given the volume of HCl gas produced (7.84 liters), we can calculate the moles of HCl using the ideal gas law. Assuming STP conditions, the moles of HCl is approximately 0.278 moles. Therefore, the moles of carbonate salt used in the reaction is 0.139 moles.

From the balanced equation M2CO3 + 2HCl -> 2MCl + CO2 + H2O, we can deduce that 2 moles of HCl react with 1 mole of carbonate salt. Since we produced 7.84 liters of HCl gas, we can use the molar volume of gases at STP (22.4 liters/mol) to calculate the moles of HCl, which is approximately equal to 0.35 moles. Therefore, the moles of carbonate salt used in the reaction is 0.175 moles.

The equation for the reaction between the carbonate salt and HCl is M2CO3 + 2HCl -> 2MCl + CO2 + H2O. Therefore, 2 moles of HCl gas are produced for every 1 mole of carbonate salt reacting. Given that 7.84 liters of HCl gas are produced, which is approximately 0.278 moles (using the ideal gas law), we can conclude that 0.139 moles of carbonate salt reacted.