Given:
"c_{C_6H_8O_7}=0.8 M = 0.8 mol\/L""m_{NaHCO_3} = 15 g"Solution:
From the reaction equation:
"3* n_{C_6H_8O_7} = n_{NaHCO_3}"
Molar mass of baking soda is:
"M_{NaHCO_3} \\approx 23 + 1 + 12 + 3 * 16 = 84 g\/mol" So,
"n_{NaHCO_3} = m_{NaHCO_3}\/M_{NaHCO_3}" And:
"n_{C_6H_8O_7} = \\frac{1}{3}*n_{NaHCO_3}" From definition of molarity:
"V_{C_6H_8O_7} = \\frac{n_{C_6H_8O_7}}{c_{C_6H_8O_7}} = \\frac{n_{NaHCO_3}}{3* c_{C_6H_8O_7}} = \\frac{m_{NaHCO_3}\/M_{NaHCO_3} }{3* c_{C_6H_8O_7}}" Calculating:
"V_{C_6H_8O_7} = \\frac{15\/84 }{3* 0.8}= \\frac{5}{84* 0.8} \\approx 0.0744 L = 74.4 mL" Answer: 74.4 mL
Comments
Leave a comment