The volume of 6.00M HBr in milliliters that would be required to react with 2.46 g of na2co3 is 7.6mL
The balanced chemical equation is:
Na2CO3(s)1 mol+2HBr(aq)2 mol→2NaBr(aq)+H2O+CO2
Moles of Na2CO3 present =2.46/106=0.023
(Molar mass =2×23+12+3×16=106)
Now, according to the equation,
1 mol of Na2CO3 requires 2 moles of HBr
0.023 mol of Na2CO3 requires moles of HBr = 2×0.023=0.046 mol
Now, 6 mole of 6M HCl is present in 1000 mL
0.046 mol of 6M HCl would be present in =(1000/6)×0.046 =7.6 mL
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