Asked by cheryl
what volume of 0.955 M HCl, in milliliters, is required to titrate 2.152g of Na2CO3 to the equivalent point?
Na2CO3(aq) + 2HCl(aq)
2NaCl(aq) + CO2(g) + H2O(1)
can this please be explained in simple step by steps? i am sick and i am having trouble understanding anything.
Na2CO3(aq) + 2HCl(aq)
2NaCl(aq) + CO2(g) + H2O(1)
can this please be explained in simple step by steps? i am sick and i am having trouble understanding anything.
Answers
Answered by
DrBob222
1. moles Na2CO3 = M x L = ?
2. Convert moles Na2CO3 to moles HCl using the coefficients in the balanced equation. You should get ?moles Na2CO3 x (2 moles HCl/1 mole Na2CO3) = ?moles HCl.
3. M HCl = moles HCl/L HCl. You know moles HCl an M HCl, solve for L HCl and convert to mL.
2. Convert moles Na2CO3 to moles HCl using the coefficients in the balanced equation. You should get ?moles Na2CO3 x (2 moles HCl/1 mole Na2CO3) = ?moles HCl.
3. M HCl = moles HCl/L HCl. You know moles HCl an M HCl, solve for L HCl and convert to mL.
Answered by
john
42.5 mL
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